Analytical geometry

Class 12 Mathematics, unit 6, book pages 160–181

Analytical geometry describes points by coordinates and lines by equations, so geometric questions become algebra. This unit uses that idea to find where lines meet, to test whether three lines pass through one point (concurrency), to find the centroid, circumcentre and orthocentre of a triangle and prove that its medians, right bisectors and altitudes are concurrent, and to find the area of a triangle from its vertices. It then shows that one homogeneous second-degree equation $a x^{2} + 2 h x y + b y^{2} = 0$ represents two straight lines through the origin, and gives the angle between them. FBISE papers test these ideas as MCQs, 4-mark short questions (area, pair of lines, one median or right bisector) and 8-mark triangle questions.

Learning outcomes (7)

Find the condition of concurrency of three straight lines.
Find the equations of altitudes, right bisectors and medians of a triangle.
Show that altitudes, right bisectors and medians of a triangle are concurrent.
Find the area of a triangular region whose vertices are given.
Recognize homogeneous linear and quadratic equations in two variables.
Investigate that a second-degree homogeneous equation in two variables represents a pair of straight lines through the origin and find the acute angle between them.
Apply concepts of analytical geometry to real-life problems such as aviation, tracking stars, distances between planets and satellites, space science and engineering.

6What analytic geometry does

Analytic geometry (also called coordinate geometry) is the branch of mathematics that joins algebra and geometry. Every point of the plane is named by an ordered pair $(x, y)$ , and every straight line or curve is described by an equation in $x$ and $y$ . A geometric question such as 'do these three roads meet at one point?' then becomes an algebra question such as 'do these three equations have a common solution?'.

It is used in physics and engineering, and in aviation, rocketry, space science, spaceflight, computer graphics, astronomy, cartography (map making) and robotics. It is also the base of newer branches of geometry: algebraic, differential, discrete and computational geometry.

This unit builds five tools on top of the Grade 11 work on straight lines:

the point where two lines meet (6.1);
a test for three lines passing through one point, called concurrency (6.2);
the special points of a triangle: centroid, circumcentre, orthocentre and incentre, and proofs that the medians, right bisectors and altitudes are concurrent (6.3);
the area of a triangle from the coordinates of its vertices (6.4);
homogeneous equations, and how one second-degree equation $a x^{2} + 2 h x y + b y^{2} = 0$ can describe two straight lines through the origin (6.5), followed by real-life applications (6.6).

Toolkit 1: distance, mid-point and section formulas

These results were proved in Grade 11. Unit 6 uses them without proof: the mid-point formula for every median and right bisector, and the distance formula in Example 2. They are collected here so that every later step can be followed.

Derivation Distance between two points

Aim. Find the distance $d$ between $P_{1} (x_{1}, y_{1})$ and $P_{2} (x_{2}, y_{2})$ .

−1

x

y

d

x_{2} - x_{1}

y_{2} - y_{1}

P_{1} (x_{1}, y_{1})

P_{2} (x_{2}, y_{2})

Q (x_{2}, y_{1})

The horizontal and vertical steps from

P_{1}

P_{2}

form a right-angled triangle with the right angle at

Q

Draw a horizontal line through $P_{1}$ and a vertical line through $P_{2}$ . They meet at $Q (x_{2}, y_{1})$ and make a right angle there, because horizontal and vertical lines are perpendicular.
The horizontal side only changes $x$ , and the vertical side only changes $y$ .
$P_{1} Q = ∣ x_{2} - x_{1} ∣, Q P_{2} = ∣ y_{2} - y_{1} ∣$
Apply Pythagoras' theorem to the right-angled triangle $P_{1} Q P_{2}$ . Squaring removes the need for the absolute value signs.
$d^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$
Take the positive square root, because a distance is never negative.
$d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$

Result

∣ P_{1} P_{2} ∣ = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

The order of the points does not matter, because $(x_{1} - x_{2})^{2} = (x_{2} - x_{1})^{2}$ . For example the distance from $(1, 1)$ to $(5, 4)$ is $16 + 9 = 5$ .

The mid-point of a segment is the average of its end points: add the $x$ -coordinates and halve, then do the same for $y$ .

M = (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})

$M$ is the mid-point of the segment joining $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ .

Derivation Section formula (internal division)

Aim. Find the point $P (x, y)$ on $A B$ with $A P : P B = m_{1} : m_{2}$ , where $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ .

−1

x

y

x - x_{1}

x_{2} - x

A (x_{1}, y_{1})

P (x, y)

B (x_{2}, y_{2})

m_{1}

m_{2}

P

divides

A B

in the ratio

m_{1} : m_{2}

. The two dashed triangles are similar, so the horizontal steps are in the same ratio.

Draw the horizontal and vertical steps from $A$ to $P$ and from $P$ to $B$ . The two small right-angled triangles are similar (their hypotenuses lie on the same line), so their horizontal sides are in the ratio $m_{1} : m_{2}$ .
$\frac{x - x _{1}}{x _{2} - x} = \frac{m _{1}}{m _{2}}$
Cross-multiply and collect the $x$ terms.
$m_{2} x - m_{2} x_{1} = m_{1} x_{2} - m_{1} x ⟹ (m_{1} + m_{2}) x = m_{1} x_{2} + m_{2} x_{1}$
Divide by $m_{1} + m_{2}$ . The vertical sides give the same result for $y$ .
$x = \frac{m _{1} x _{2} + m _{2} x _{1}}{m _{1} + m _{2}}, y = \frac{m _{1} y _{2} + m _{2} y _{1}}{m _{1} + m _{2}}$

Result

P = (\frac{m _{1} x _{2} + m _{2} x _{1}}{m _{1} + m _{2}}, \frac{m _{1} y _{2} + m _{2} y _{1}}{m _{1} + m _{2}})

$P$ is a weighted average of $A$ and $B$ . With $m_{1} = m_{2}$ it becomes the mid-point formula. This formula gives the centroid in section 6.3, because the centroid divides each median in the ratio $2 : 1$ .

Point dividing a segment in a given ratio

Find the point that divides the segment joining

A (1, 2)

and

B (4, 5)

internally in the ratio

2 : 1

Here $m_{1} = 2$ , $m_{2} = 1$ , $(x_{1}, y_{1}) = (1, 2)$ and $(x_{2}, y_{2}) = (4, 5)$ . Use the section formula.
$x = \frac{2 ( 4 ) + 1 ( 1 )}{2 + 1} = \frac{9}{3} = 3, y = \frac{2 ( 5 ) + 1 ( 2 )}{2 + 1} = \frac{12}{3} = 4$
Check: $A P = 2^{2} + 2^{2} = 22$ and $P B = 1^{2} + 1^{2} = 2$ , so $A P : P B = 2 : 1$ .

−1

x

y

A (1, 2)

P (3, 4)

B (4, 5)

2

1

P

is two parts from

A

and one part from

B

, so it lies nearer to

B

P (3, 4)

Collinear points lie on one straight line. With distances, $A$ , $B$ , $C$ are collinear when the largest distance equals the sum of the other two, for example $A C = A B + B C$ . Example 2 of the book uses this test. The area test of section 6.4 (area $= 0$ ) is usually quicker.

Toolkit 2: slope and the forms of the equation of a lineOften examined

The inclination \alpha of a line is the angle it makes with the positive x-axis, measured anticlockwise, with 0^\circ\le\alpha<180^\circ. The slope (gradient) m is the rise per unit run.

m=\tan\alpha=\dfrac{y_2-y_1}{x_2-x_1}

(x_1,y_1) and (x_2,y_2) are any two points of the line, with x_1\ne x_2.

m>0: the line rises to the right; m<0: it falls to the right.
m=0: horizontal line y=k. A vertical line x=k has no slope (the book writes m=\infty).
Parallel lines have equal slopes: m_1=m_2.
Perpendicular lines have slopes whose product is -1: m_1m_2=-1, that is m_2=-\dfrac{1}{m_1}. This rule gives the slope of every altitude and right bisector in this unit.

Forms of the equation of a straight line
Form	Equation	Use it when you know
Slope-intercept	y=mx+c	slope m and y-intercept c
Point-slope	y-y_1=m(x-x_1)	slope and one point (altitudes, right bisectors)
Two-point	\dfrac{y-y_1}{y_2-y_1}=\dfrac{x-x_1}{x_2-x_1}	two points (medians)
Two-intercept	\dfrac{x}{a}+\dfrac{y}{b}=1	x-intercept a and y-intercept b
Normal	x\cos\alpha+y\sin\alpha=p	length p of the perpendicular from O and its angle \alpha
General	ax+by+c=0	every line; slope -\dfrac ab, intercepts -\dfrac ca and -\dfrac cb
Axis-parallel	x=k or y=k	vertical or horizontal lines

Changing the form of a line

Write 2x+3y-12=0 in slope-intercept form and in two-intercept form.

Slope-intercept form. Keep y alone on the left side.
3y=-2x+12\ \Longrightarrow\ y=-\tfrac23x+4
Read off the slope and the y-intercept.
m=-\tfrac23,\qquad c=4
Two-intercept form. Move the constant to the right side, then divide by it so that the right side becomes 1.
2x+3y=12\ \Longrightarrow\ \dfrac{x}{6}+\dfrac{y}{4}=1
So the x-intercept is 6 and the y-intercept is 4.

The line cuts the x-axis at 6 and the y-axis at 4; it falls 2 units for every 3 units to the right, so m=-\tfrac23.

y=-\tfrac23x+4,\qquad \dfrac{x}{6}+\dfrac{y}{4}=1

Parallel and perpendicular lines through a point. A line parallel to ax+by+c=0 has the same a and b, so it is ax+by+k=0. A line perpendicular to it swaps a and b and changes one sign, so it is bx-ay+k=0. Find k by substituting the given point.

Line through a point perpendicular to a given line

Find the line through (1,2) perpendicular to 3x+4y-5=0.

A perpendicular line has the form 4x-3y+k=0. Check the slopes: -\tfrac34 and \tfrac43 multiply to -1.
It passes through (1,2), so substitute to find k.
4(1)-3(2)+k=0\ \Longrightarrow\ k=2

The required line passes through (1,2) and crosses the given line at right angles.

4x-3y+2=0

Toolkit 3: angle between lines, distance from a line, position of a point

Derivation Angle between two lines

Aim. Find the angle \theta from line l_1 (slope m_1) to line l_2 (slope m_2).

The exterior angle \alpha_2 of the triangle equals \alpha_1+\theta, so \theta=\alpha_2-\alpha_1.

Let the inclinations be \alpha_1 and \alpha_2, so m_1=\tan\alpha_1 and m_2=\tan\alpha_2. In the triangle formed with the x-axis, the exterior angle \alpha_2 equals the sum of the two opposite interior angles \alpha_1 and \theta.
\theta=\alpha_2-\alpha_1
Take the tangent of both sides and use the subtraction formula for \tan.
\tan\theta=\dfrac{\tan\alpha_2-\tan\alpha_1}{1+\tan\alpha_1\tan\alpha_2}
Replace the tangents by the slopes.
\tan\theta=\dfrac{m_2-m_1}{1+m_1m_2}
Two crossing lines make two angles, \theta and 180^\circ-\theta, whose tangents differ only in sign. The acute angle has a positive tangent, so take the absolute value.
\tan\theta_{\text{acute}}=\left|\dfrac{m_2-m_1}{1+m_1m_2}\right|

Result\tan\theta=\left|\dfrac{m_2-m_1}{1+m_1m_2}\right|

If m_1=m_2 then \tan\theta=0: the lines are parallel. If 1+m_1m_2=0 the denominator is zero, so \tan\theta has no finite value and \theta=90^\circ: the lines are perpendicular. This is why m_1m_2=-1 is the perpendicular condition.

Acute angle between two lines

Find the acute angle between y=3x-2 and y=-x+2.

Read the slopes from y=mx+c.
m_1=3,\qquad m_2=-1
Substitute in the angle formula.
\tan\theta=\left|\dfrac{-1-3}{1+(3)(-1)}\right|=\left|\dfrac{-4}{-2}\right|=2
Use the inverse tangent.
\theta=\tan^{-1}2=63.43^\circ

\theta\approx63.43^\circ

Derivation Distance of a point from a line

Aim. Find the perpendicular distance d of P(x_1,y_1) from the line ax+by+c=0.

PN is perpendicular to the line; its length d is the distance of P from the line.

The line has slope -\dfrac ab. The direction (a,b) has slope \dfrac ba, and \left(-\dfrac ab\right)\left(\dfrac ba\right)=-1, so (a,b) is perpendicular to the line (it is called a normal direction).
Let N be the foot of the perpendicular from P. Moving from P along the normal direction reaches N, so for some number t:
N=(x_1-ta,\ y_1-tb)
N lies on the line, so its coordinates satisfy the equation. Solve for t.
a(x_1-ta)+b(y_1-tb)+c=0\ \Longrightarrow\ t=\dfrac{ax_1+by_1+c}{a^2+b^2}
The distance is the length of PN, which is |t| times the length of (a,b).
d=|t|\sqrt{a^2+b^2}=\dfrac{|ax_1+by_1+c|}{a^2+b^2}\sqrt{a^2+b^2}

Resultd=\dfrac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}

Substitute the point into the left side of the equation, take the absolute value and divide by \sqrt{a^2+b^2}. If the value is 0 the point is on the line.

Distance of a point from a line

Find the distance of (-1,2) from the line 5x-12y+3=0.

Here a=5, b=-12, c=3 and (x_1,y_1)=(-1,2).
d=\dfrac{|5(-1)-12(2)+3|}{\sqrt{5^2+(-12)^2}}=\dfrac{|-26|}{13}=2

d=2\ \text{units}

The distance between two parallel lines ax+by+c_1=0 and ax+by+c_2=0 (same a and b) is

d=\dfrac{|c_1-c_2|}{\sqrt{a^2+b^2}}

Position of a point relative to a line

Substitute the point (x_1,y_1) into the left side ax+by+c. The value is 0 exactly when the point is on the line. All points on one side of the line give values of one sign, and points on the other side give the opposite sign.

Why the sign tells the side

First make the coefficient of y positive (b>0); multiply the equation by -1 if necessary. The line is then y=-\dfrac{ax+c}{b}.
The point is above the line when its y is larger than the line's y at the same x.
y_1>-\dfrac{ax_1+c}{b}
Multiply by b>0 (the inequality keeps its direction) and bring every term to the left.
ax_1+by_1+c>0

With b>0: ax_1+by_1+c>0 means above the line, <0 means below, =0 means on the line.
Two points are on the same side of a line when they give values of the same sign, whatever the sign of b.
The origin gives the value c, so a point is on the origin side when its value has the same sign as c.

With the coefficient of y positive, points above the line give a positive value and points below give a negative value. The origin gives -3, so it is below.

Position of a point

Where does the point (2,-1) lie relative to the line x+2y-3=0?

The coefficient of y is b=2>0, so the sign rule can be used directly.
Substitute the point.
2+2(-1)-3=-3<0
The value is negative, so the point is below the line. The origin gives -3 too, so the point is on the same side as the origin.

(2,-1)\ \text{lies below the line}

6.1Point of intersection of two straight linesOften examined

Two lines that are not parallel meet at exactly one point. That point lies on both lines, so its coordinates satisfy both equations at the same time. Finding it means solving two simultaneous linear equations.

l_1:\ a_1x+b_1y+c_1=0\qquad\qquad l_2:\ a_2x+b_2y+c_2=0

Derivation The cross-multiplication rule

Aim. Find the point of intersection P(x_1,y_1) of l_1 and l_2.

P lies on both lines, so it satisfies both equations.
a_1x_1+b_1y_1+c_1=0\quad(3)\qquad a_2x_1+b_2y_1+c_2=0\quad(4)
Eliminate y_1: multiply (3) by b_2 and (4) by b_1, then subtract.
(a_1b_2-a_2b_1)x_1+(b_2c_1-b_1c_2)=0\ \Longrightarrow\ x_1=\dfrac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}
Eliminate x_1: multiply (3) by a_2 and (4) by a_1, then subtract.
(a_2b_1-a_1b_2)y_1+(a_2c_1-a_1c_2)=0\ \Longrightarrow\ y_1=\dfrac{a_2c_1-a_1c_2}{a_1b_2-a_2b_1}
Both answers fit one memory pattern. Write the coefficients in the order b,\ c,\ a,\ b and cross-multiply each pair of neighbouring columns (down-right product minus up-right product).
\begin{array}{cccc} b_1 & c_1 & a_1 & b_1\\ b_2 & c_2 & a_2 & b_2\end{array}\qquad \dfrac{x_1}{b_1c_2-b_2c_1}=\dfrac{y_1}{a_2c_1-a_1c_2}=\dfrac{1}{a_1b_2-a_2b_1}

ResultP(x_1,y_1)=\left(\dfrac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1},\ \dfrac{a_2c_1-a_1c_2}{a_1b_2-a_2b_1}\right),\qquad a_1b_2-a_2b_1\ne0

Everything depends on the denominator a_1b_2-a_2b_1. If it is not zero the lines meet at exactly one point. If it is zero the lines are parallel (or they are the same line) and there is no single point of intersection.

Two linear equations and the lines they represent
Condition	The two lines	Solutions of the system
\dfrac{a_1}{a_2}\ne\dfrac{b_1}{b_2}	intersect at one point	exactly one (unique) solution
\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\ne\dfrac{c_1}{c_2}	are parallel and distinct	no solution
\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}	are the same (coincident) line	infinitely many solutions

Intersection by the cross-multiplication rule

Find the point of intersection of 3x-2y+1=0 and x+y-3=0.

Here a_1=3,\ b_1=-2,\ c_1=1 and a_2=1,\ b_2=1,\ c_2=-3. First check the denominator: it is not zero, so the lines do meet.
a_1b_2-a_2b_1=3(1)-1(-2)=5
Numerators from the pattern b,\ c,\ a,\ b.
\begin{aligned}b_1c_2-b_2c_1&=(-2)(-3)-(1)(1)\\ &=5\\ a_2c_1-a_1c_2&=(1)(1)-(3)(-3)\\ &=10\end{aligned}
Divide by the denominator.
x=\dfrac{5}{5}=1,\qquad y=\dfrac{10}{5}=2
Check in both equations: 3-4+1=0 and 1+2-3=0.

The two lines meet at exactly one point, (1,2).

(1,\ 2)

6.2Concurrent lines and the condition of concurrencyOften examined

Concurrent lines (6.2.1): Three or more lines that pass through one common point. When a third line passes through the point of intersection of the first two lines, the three lines are concurrent.
Point of concurrency (6.2.2): The common point where three or more concurrent lines intersect.

Two lines that are not parallel always meet. Three lines usually do not: each pair meets at a different point and the three points form a small triangle. The lines are concurrent only when the point where two of them meet also lies on the third, that is, it satisfies the equation of the third line.

All three lines pass through (4,6): they are concurrent.

(a) Determinant method

Derivation Condition of concurrency of three lines

Aim. Find a condition on the coefficients for the lines a_1x+b_1y+c_1=0, a_2x+b_2y+c_2=0, a_3x+b_3y+c_3=0 to pass through one point.

Suppose the three lines are concurrent at (x_0,y_0). Then all three equations are true at the same time.
\begin{aligned} a_1x_0+b_1y_0+c_1(1)&=0\\ a_2x_0+b_2y_0+c_2(1)&=0\\ a_3x_0+b_3y_0+c_3(1)&=0\end{aligned}
Write the three equations as one matrix equation. The constant term is treated as c\times1, so the column of unknowns is (x_0,\ y_0,\ 1).
\begin{bmatrix} a_1&b_1&c_1\\ a_2&b_2&c_2\\ a_3&b_3&c_3\end{bmatrix}\begin{bmatrix} x_0\\ y_0\\ 1\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}\qquad(4)
(4) is a homogeneous system. It has a non-zero solution, because the third entry of (x_0,y_0,1) is 1. A homogeneous system with a non-singular coefficient matrix has only the zero (trivial) solution, so the coefficient matrix must be singular.
A singular matrix has determinant zero.
\begin{vmatrix} a_1&b_1&c_1\\ a_2&b_2&c_2\\ a_3&b_3&c_3\end{vmatrix}=0
Conversely, if the determinant is zero and two of the lines intersect, the third line passes through their point of intersection, so the lines are concurrent.

Result\begin{vmatrix} a_1&b_1&c_1\\ a_2&b_2&c_2\\ a_3&b_3&c_3\end{vmatrix}=0

Use the condition two ways: to test given lines (evaluate the determinant) and to find an unknown coefficient (set the determinant equal to zero and solve).

Example 1: finding an unknown coefficient

Find the value of k if the lines 3x+y-3=0, 5x+ky-3=0 and 3x-y-2=0 are concurrent.

Write the determinant of coefficients.
D=\begin{vmatrix} 3&1&-3\\ 5&k&-3\\ 3&-1&-2\end{vmatrix}
Expand along the first row.
D=3\begin{vmatrix} k&-3\\ -1&-2\end{vmatrix}-1\begin{vmatrix} 5&-3\\ 3&-2\end{vmatrix}+(-3)\begin{vmatrix} 5&k\\ 3&-1\end{vmatrix}
Evaluate the 2\times2 determinants.
D=3(-2k-3)-1(-10+9)-3(-5-3k)=-6k-9+1+15+9k=3k+7
For concurrent lines D=0.
3k+7=0\ \Longrightarrow\ k=-\dfrac{7}{3}
Check: the first and third lines meet where 6x-5=0, at \left(\tfrac56,\tfrac12\right). Then 5\left(\tfrac56\right)-\tfrac73\left(\tfrac12\right)-3=\tfrac{25}{6}-\tfrac{7}{6}-\tfrac{18}{6}=0, so the second line passes through it as well.

With k=-\tfrac73 the three lines pass through \left(\tfrac56,\tfrac12\right).

k=-\dfrac{7}{3}

Check point: lines that are not concurrent

Check whether the lines 3x+4y-7=0, 2x-3y+5=0 and 3x-5y+8=0 are concurrent or not.

Evaluate the determinant of coefficients, expanding along the first row.
\begin{vmatrix} 3&4&-7\\ 2&-3&5\\ 3&-5&8\end{vmatrix}=3(-24+25)-4(16-15)-7(-10+9)=3-4+7=6
The determinant is 6\ne0, so the lines are not concurrent.

Zoomed in: the lines cross at three different points, \left(\tfrac1{17},\tfrac{29}{17}\right), \left(\tfrac19,\tfrac53\right) and (-1,1), forming a thin triangle. They are not concurrent.

\text{Not concurrent } (D=6\ne0)

(b) Direct method

Find the point of intersection of two of the lines, then check whether it satisfies the equation of the third line.

Direct method

Show that the lines 4x-2y-4=0 (1), y=x+2 (2) and 2x+3y=26 (3) are concurrent.

Step 1. Substitute y from (2) into (1) to find where (1) and (2) meet.
4x-2(x+2)-4=0\ \Longrightarrow\ 2x-8=0\ \Longrightarrow\ x=4
Put x=4 in (2).
y=4+2=6\qquad\text{so (1) and (2) meet at }(4,6)
Step 2. Substitute (4,6) in the left side of (3).
2(4)+3(6)=8+18=26
The third line also passes through (4,6), so the three lines are concurrent at (4,6).

\text{Concurrent at } (4,\ 6)

6.3Concurrent lines in a triangleOften examined

Every triangle has four sets of three special lines, and each set is concurrent. The four points of concurrency are called the centres of the triangle.

Median: The segment joining a vertex to the mid-point of the opposite side.
Altitude: The perpendicular from a vertex to the opposite side (extended if necessary).
Right bisector: The line through the mid-point of a side, perpendicular to that side (also called the perpendicular bisector).
Angle bisector: The line through a vertex that divides the angle at that vertex into two equal angles.
Incentre: The point of intersection of the three angle bisectors of a triangle.
Circumcentre: The point of intersection of the three perpendicular (right) bisectors of a triangle.
Centroid: The point of intersection of the three medians of a triangle.
Orthocentre: The point of intersection of the three altitudes of a triangle.

The four centres of the same triangle. Accent lines: angle bisectors, right bisectors, medians and altitudes.

The four centres of a triangle
Lines	Meet at	Useful property
Angle bisectors	Incentre I	Same distance from all three sides; centre of the circle inside the triangle touching all sides. Always inside.
Right bisectors	Circumcentre O	Same distance from all three vertices; centre of the circle through A, B, C. Inside an acute triangle, at the mid-point of the hypotenuse of a right triangle, outside an obtuse triangle.
Medians	Centroid G	Divides each median in the ratio 2:1 from the vertex; G=\left(\tfrac{x_1+x_2+x_3}{3},\tfrac{y_1+y_2+y_3}{3}\right). Always inside.
Altitudes	Orthocentre H	Inside an acute triangle, at the right-angle vertex of a right triangle, outside an obtuse triangle.

How to find the equation of each line

**Median from A:** find the mid-point D of BC, then use the two-point form through A and D.
**Altitude from A:** find the slope m of BC; the altitude has slope -\dfrac1m; use the point-slope form through A.
**Right bisector of BC:** find the mid-point D of BC and the slope m of BC; the bisector has slope -\dfrac1m; use the point-slope form through D.
Special cases: if BC is horizontal (m=0), the altitude and the right bisector are vertical lines x=k; if BC is vertical, they are horizontal lines y=k.
Point of concurrency: solve any two of the three equations, then check that the point satisfies the third.

Derivation Coordinates of the centroid

Aim. Find the point G where the medians of the triangle with vertices A(x_1,y_1), B(x_2,y_2), C(x_3,y_3) meet.

The centroid G is two-thirds of the way from A to the mid-point D of BC.

The median from A ends at the mid-point D of BC.
D=\left(\dfrac{x_2+x_3}{2},\ \dfrac{y_2+y_3}{2}\right)
Take the point that divides AD in the ratio 2:1 and use the section formula with m_1=2, m_2=1.
x=\dfrac{2\cdot\frac{x_2+x_3}{2}+1\cdot x_1}{2+1}=\dfrac{x_1+x_2+x_3}{3},\qquad y=\dfrac{y_1+y_2+y_3}{3}
This answer is symmetric in the three vertices. Doing the same for the medians from B and from C gives exactly the same point, so this one point lies on all three medians.

ResultG=\left(\dfrac{x_1+x_2+x_3}{3},\ \dfrac{y_1+y_2+y_3}{3}\right)

This is a second, shorter proof that the medians are concurrent, and it shows that the centroid divides every median in the ratio 2:1 from the vertex. Use it to check a centroid found by solving equations.

6.3Theorem 6.1: the altitudes of a triangle are concurrentOften examined

The proof finds the equation of one altitude in general form, gets the other two by symmetry, and then shows that the determinant of their coefficients is zero.

Derivation Theorem 6.1

Aim. Prove that the altitudes of a triangle ABC with vertices A(x_1,y_1), B(x_2,y_2), C(x_3,y_3) are concurrent.

The altitudes AD\perp BC, BE\perp CA and CF\perp AB meet at one point O, the orthocentre.

AD, BE and CF are the altitudes to BC, CA and AB. First find the slope of BC; AD is perpendicular to BC, so its slope is the negative reciprocal.
m_{BC}=\dfrac{y_3-y_2}{x_3-x_2},\qquad m_{AD}=-\dfrac{x_3-x_2}{y_3-y_2}
AD passes through A(x_1,y_1). Use the point-slope form.
y-y_1=-\dfrac{x_3-x_2}{y_3-y_2}\,(x-x_1)
Multiply both sides by (y_3-y_2) to clear the fraction.
(y-y_1)(y_3-y_2)=-(x_3-x_2)(x-x_1)
Expand, move every term to the left side and collect the x and y terms. This is the equation of AD in general form.
(x_3-x_2)x+(y_3-y_2)y-x_1(x_3-x_2)-y_1(y_3-y_2)=0\qquad(1)
'By symmetry': the altitude BE is found in exactly the same way, so its equation comes from (1) by moving every subscript one step round the cycle 1\to2\to3\to1. Doing this once more gives CF.
\begin{aligned}(x_1-x_3)x+(y_1-y_3)y-x_2(x_1-x_3)-y_2(y_1-y_3)&=0\qquad(2)\\ (x_2-x_1)x+(y_2-y_1)y-x_3(x_2-x_1)-y_3(y_2-y_1)&=0\qquad(3)\end{aligned}
Write the determinant of the coefficients of (1), (2) and (3).
\Delta=\begin{vmatrix} x_3-x_2 & y_3-y_2 & \begin{array}{l}-x_1(x_3-x_2)\\-y_1(y_3-y_2)\end{array}\\[12pt] x_1-x_3 & y_1-y_3 & \begin{array}{l}-x_2(x_1-x_3)\\-y_2(y_1-y_3)\end{array}\\[12pt] x_2-x_1 & y_2-y_1 & \begin{array}{l}-x_3(x_2-x_1)\\-y_3(y_2-y_1)\end{array}\end{vmatrix}
Add R_2 and R_3 to R_1 (this does not change the value of a determinant). In the first two columns the terms cancel in pairs.
\begin{aligned}(x_3-x_2)+(x_1-x_3)+(x_2-x_1)&=0\\ (y_3-y_2)+(y_1-y_3)+(y_2-y_1)&=0\end{aligned}
In the third column every product x_ix_j and y_iy_j appears once with a plus sign and once with a minus sign.
\begin{aligned}&-x_1x_3+x_1x_2-x_2x_1+x_2x_3-x_3x_2+x_3x_1\\ &-y_1y_3+y_1y_2-y_2y_1+y_2y_3-y_3y_2+y_3y_1=0\end{aligned}
So R_1 becomes a row of zeros, and a determinant with a zero row is zero.
\Delta=\begin{vmatrix} 0 & 0 & 0\\[6pt] x_1-x_3 & y_1-y_3 & \begin{array}{l}-x_2(x_1-x_3)\\-y_2(y_1-y_3)\end{array}\\[12pt] x_2-x_1 & y_2-y_1 & \begin{array}{l}-x_3(x_2-x_1)\\-y_3(y_2-y_1)\end{array}\end{vmatrix}=0
The altitudes are perpendicular to three sides that are not parallel, so no two altitudes are parallel. By the condition of concurrency the three altitudes pass through one point.

Result\Delta=0\ \Longrightarrow\ \text{the altitudes } AD,\ BE,\ CF \text{ are concurrent}

Every triangle has an orthocentre. To locate it, two altitudes are enough; the third one is only a check.

6.3Theorem 6.2: the right bisectors of a triangle are concurrentOften examined

Derivation Theorem 6.2

Aim. Prove that the right bisectors of the sides of a triangle ABC with vertices A(x_1,y_1), B(x_2,y_2), C(x_3,y_3) are concurrent.

D, E, F are the mid-points of the sides. The right bisectors through them meet at O, the circumcentre.

OD, OE, OF are the right bisectors of BC, CA, AB. The right bisector of BC passes through the mid-point D and is perpendicular to BC.
D=\left(\dfrac{x_2+x_3}{2},\ \dfrac{y_2+y_3}{2}\right),\qquad m_{OD}=-\dfrac{x_3-x_2}{y_3-y_2}
Use the point-slope form through D.
y-\dfrac{y_2+y_3}{2}=-\dfrac{x_3-x_2}{y_3-y_2}\left(x-\dfrac{x_2+x_3}{2}\right)
Multiply by (y_3-y_2) and bring every term to the left.
(x_3-x_2)x+(y_3-y_2)y-\tfrac12(x_3-x_2)(x_3+x_2)-\tfrac12(y_3-y_2)(y_3+y_2)=0
Use (p-q)(p+q)=p^2-q^2. This is the equation of the right bisector of BC.
(x_3-x_2)x+(y_3-y_2)y-\tfrac12(x_3^2-x_2^2)-\tfrac12(y_3^2-y_2^2)=0\qquad(1)
By symmetry (move the subscripts round 1\to2\to3\to1) the right bisectors of CA and AB are:
\begin{aligned}(x_1-x_3)x+(y_1-y_3)y-\tfrac12(x_1^2-x_3^2)-\tfrac12(y_1^2-y_3^2)&=0\qquad(2)\\ (x_2-x_1)x+(y_2-y_1)y-\tfrac12(x_2^2-x_1^2)-\tfrac12(y_2^2-y_1^2)&=0\qquad(3)\end{aligned}
Write the determinant of coefficients.
\Delta=\begin{vmatrix} x_3-x_2 & y_3-y_2 & \begin{array}{l}-\tfrac12(x_3^2-x_2^2)\\[2pt]-\tfrac12(y_3^2-y_2^2)\end{array}\\[14pt] x_1-x_3 & y_1-y_3 & \begin{array}{l}-\tfrac12(x_1^2-x_3^2)\\[2pt]-\tfrac12(y_1^2-y_3^2)\end{array}\\[14pt] x_2-x_1 & y_2-y_1 & \begin{array}{l}-\tfrac12(x_2^2-x_1^2)\\[2pt]-\tfrac12(y_2^2-y_1^2)\end{array}\end{vmatrix}
Add R_2 and R_3 to R_1. The first two columns cancel as in Theorem 6.1. In the third column each square appears once with + and once with -.
-\tfrac12\left[(x_3^2-x_2^2)+(x_1^2-x_3^2)+(x_2^2-x_1^2)\right]-\tfrac12\left[(y_3^2-y_2^2)+(y_1^2-y_3^2)+(y_2^2-y_1^2)\right]=0
R_1 is a zero row, so \Delta=0. The right bisectors are perpendicular to non-parallel sides, so no two of them are parallel. Hence they are concurrent.
\Delta=\begin{vmatrix} 0&0&0\\[6pt] x_1-x_3 & y_1-y_3 & \begin{array}{l}-\tfrac12(x_1^2-x_3^2)\\[2pt]-\tfrac12(y_1^2-y_3^2)\end{array}\\[14pt] x_2-x_1 & y_2-y_1 & \begin{array}{l}-\tfrac12(x_2^2-x_1^2)\\[2pt]-\tfrac12(y_2^2-y_1^2)\end{array}\end{vmatrix}=0

Result\Delta=0\ \Longrightarrow\ \text{the right bisectors } OD,\ OE,\ OF \text{ are concurrent}

The point of concurrency O is the circumcentre: it is equally far from A, B and C, so a circle with centre O passes through all three vertices.

6.3Theorem 6.3: the medians of a triangle are concurrentOften examined

Derivation Theorem 6.3

Aim. Prove that the medians of a triangle ABC with vertices A(x_1,y_1), B(x_2,y_2), C(x_3,y_3) are concurrent.

The medians AD, BE, CF join each vertex to the mid-point of the opposite side and meet at O, the centroid.

AD, BE, CF are the medians. The median AD joins A to the mid-point D of BC.
D=\left(\dfrac{x_2+x_3}{2},\ \dfrac{y_2+y_3}{2}\right)
Use the two-point form through A and D.
\dfrac{y-y_1}{\frac{y_2+y_3}{2}-y_1}=\dfrac{x-x_1}{\frac{x_2+x_3}{2}-x_1}
To keep the algebra readable, write p_1=\dfrac{y_2+y_3}{2}-y_1 and q_1=\dfrac{x_2+x_3}{2}-x_1 (the book writes them out in full). Cross-multiply.
q_1(y-y_1)=p_1(x-x_1)
Bring every term to the left. This is the median AD in general form.
p_1x-q_1y-p_1x_1+q_1y_1=0\qquad(1)
By symmetry the medians BE and CF are
p_2x-q_2y-p_2x_2+q_2y_2=0\ (2),\qquad p_3x-q_3y-p_3x_3+q_3y_3=0\ (3)
Here p_2=\frac{y_1+y_3}{2}-y_2, q_2=\frac{x_1+x_3}{2}-x_2, p_3=\frac{y_1+y_2}{2}-y_3, q_3=\frac{x_1+x_2}{2}-x_3.
Write the determinant of coefficients.
\Delta=\begin{vmatrix} p_1 & -q_1 & -p_1x_1+q_1y_1\\ p_2 & -q_2 & -p_2x_2+q_2y_2\\ p_3 & -q_3 & -p_3x_3+q_3y_3\end{vmatrix}
Add R_2 and R_3 to R_1. In the first column each y appears twice with +\tfrac12 and once with -1; the second column is the same with x.
p_1+p_2+p_3=(y_1+y_2+y_3)-(y_1+y_2+y_3)=0,\qquad q_1+q_2+q_3=0
In the third column the x_iy_i terms cancel inside each entry, leaving half-products. Both brackets contain the same six products x_iy_j with i\ne j, so the sum is zero.
-\tfrac12\left[x_1(y_2+y_3)+x_2(y_1+y_3)+x_3(y_1+y_2)\right]+\tfrac12\left[y_1(x_2+x_3)+y_2(x_1+x_3)+y_3(x_1+x_2)\right]=0
R_1 is a zero row, so \Delta=0. Two medians cannot be parallel (they meet inside the triangle), so the medians are concurrent.

Result\Delta=0\ \Longrightarrow\ \text{the medians } AD,\ BE,\ CF \text{ are concurrent}

The point of concurrency is the centroid G=\left(\tfrac{x_1+x_2+x_3}{3},\tfrac{y_1+y_2+y_3}{3}\right). The section on the centroid gives a shorter proof with the section formula.

6.3Worked examples: centroid, circumcentre and orthocentreOften examined

Example 2

Find (i) the circumcentre, (ii) the centroid and (iii) the orthocentre of the triangle ABC with A(0,0), B(6,0) and C(0,6). Also prove that the circumcentre, centroid and orthocentre are collinear.

(i) Circumcentre

Mid-points of the three sides.
AB:\ (3,0),\qquad AC:\ (0,3),\qquad BC:\ (3,3)
Slopes of the sides.
m_{AB}=\dfrac{0-0}{6-0}=0,\qquad m_{AC}=\dfrac{6-0}{0-0}\ (\text{undefined}),\qquad m_{BC}=\dfrac{6-0}{0-6}=-1
AB is horizontal, so its right bisector is vertical through (3,0).
x-3=0\qquad(1)
AC is vertical, so its right bisector is horizontal through (0,3).
y-3=0\qquad(2)
The right bisector of BC has slope -\dfrac{1}{-1}=1 and passes through (3,3).
y-3=1(x-3)\ \Longrightarrow\ y-x=0\qquad(3)
Solve (1) and (2): x=3, y=3. This point also satisfies (3).
\text{Circumcentre } S(3,3)

(ii) Centroid

Median from A(0,0) to the mid-point (3,3) of BC (two-point form).
\dfrac{y-0}{3-0}=\dfrac{x-0}{3-0}\ \Longrightarrow\ y=x\qquad(4)
Median from B(6,0) to the mid-point (0,3) of AC.
\dfrac{y-0}{3-0}=\dfrac{x-6}{0-6}\ \Longrightarrow\ -6y=3x-18\ \Longrightarrow\ x+2y-6=0\qquad(5)
Median from C(0,6) to the mid-point (3,0) of AB.
\dfrac{y-6}{0-6}=\dfrac{x-0}{3-0}\ \Longrightarrow\ 3y-18=-6x\ \Longrightarrow\ 2x+y-6=0\qquad(6)
Put y=x from (4) into (5): 3x=6, so x=2, y=2. Check with the formula: \left(\tfrac{0+6+0}{3},\tfrac{0+0+6}{3}\right)=(2,2).
\text{Centroid } G(2,2)

(iii) Orthocentre

The angle at A is a right angle (AB is along the x-axis and AC along the y-axis). So the sides AB and AC are themselves altitudes.
y=0\ (7),\qquad x=0\ (8)
The third altitude from A is perpendicular to BC (slope -1), so its slope is 1.
y-0=1(x-0)\ \Longrightarrow\ y-x=0\qquad(9)
Solve (7) and (8): x=0, y=0, which also satisfies (9).
\text{Orthocentre } H(0,0)

Collinearity

Find the three distances between H(0,0), G(2,2) and S(3,3).
\begin{aligned}HG&=\sqrt{2^2+2^2}=2\sqrt2\\ GS&=\sqrt{1^2+1^2}=\sqrt2\\ HS&=\sqrt{3^2+3^2}=3\sqrt2\end{aligned}
The largest distance is the sum of the other two, so the three points lie on one line (here the line y=x).
HG+GS=2\sqrt2+\sqrt2=3\sqrt2=HS

Orthocentre H (at the vertex A), centroid G and circumcentre S all lie on y=x, with HG:GS=2:1. S is the mid-point of the hypotenuse BC.

S(3,3),\quad G(2,2),\quad H(0,0)\ \text{are collinear}

Example 3

The points P(-1,2), Q(3,-2) and R(6,3) are the vertices of a triangle PQR. Show that the altitudes, right bisectors and medians of the triangle are concurrent.

Let (x_1,y_1)=(-1,2), (x_2,y_2)=(3,-2) and (x_3,y_3)=(6,3), and substitute into the three general determinants of Theorems 6.1, 6.2 and 6.3.

Altitudes

Substitute into the determinant of Theorem 6.1. For example the first row is x_3-x_2=3, y_3-y_2=5 and -x_1(x_3-x_2)-y_1(y_3-y_2)=1(3)-2(5)=-7.
\begin{vmatrix} 6-3 & 3+2 & 1(6-3)-2(3+2)\\ -1-6 & 2-3 & -3(-1-6)+2(2-3)\\ 3+1 & -2-2 & -6(3+1)-3(-2-2)\end{vmatrix}=\begin{vmatrix} 3&5&-7\\ -7&-1&19\\ 4&-4&-12\end{vmatrix}
Add R_3 to R_2, then multiply R_2 by -1. Two rows become identical, so the determinant is zero.
\begin{vmatrix} 3&5&-7\\ -3&-5&7\\ 4&-4&-12\end{vmatrix}=-\begin{vmatrix} 3&5&-7\\ 3&5&-7\\ 4&-4&-12\end{vmatrix}=0
Hence the altitudes are concurrent.

Right bisectors

Substitute into the determinant of Theorem 6.2. For example the first row's last entry is -\tfrac12(36-9)-\tfrac12(9-4)=-13.5-2.5=-16.
\begin{vmatrix} 3&5&-16\\ -7&-1&20\\ 4&-4&-4\end{vmatrix}
Add R_3 to R_2, then add the new R_2 to R_1. R_1 becomes a zero row.
\begin{vmatrix} 3&5&-16\\ -3&-5&16\\ 4&-4&-4\end{vmatrix}=\begin{vmatrix} 0&0&0\\ -3&-5&16\\ 4&-4&-4\end{vmatrix}=0
Hence the right bisectors are concurrent.

Medians

Substitute into the determinant of Theorem 6.3. First row: p_1=\tfrac{-2+3}{2}-2=-1.5, -q_1=-\left(\tfrac{3+6}{2}+1\right)=-5.5, last entry -(-1.5)(-1)+(5.5)(2)=9.5.
\begin{vmatrix} -1.5&-5.5&9.5\\ 4.5&0.5&-12.5\\ -3&5&3\end{vmatrix}
Add R_3 to R_2, then add the new R_2 to R_1.
\begin{vmatrix} -1.5&-5.5&9.5\\ 1.5&5.5&-9.5\\ -3&5&3\end{vmatrix}=\begin{vmatrix} 0&0&0\\ 1.5&5.5&-9.5\\ -3&5&3\end{vmatrix}=0
Hence the medians are concurrent.

For completeness, solving pairs of the equations gives the three points of concurrency: orthocentre H\left(\tfrac{11}{4},-\tfrac14\right), circumcentre O\left(\tfrac{21}{8},\tfrac{13}{8}\right) and centroid G\left(\tfrac83,1\right). The centroid agrees with \left(\tfrac{-1+3+6}{3},\tfrac{2-2+3}{3}\right).

Orthocentre H\left(\tfrac{11}4,-\tfrac14\right), centroid G\left(\tfrac83,1\right) and circumcentre O\left(\tfrac{21}8,\tfrac{13}8\right) of triangle PQR.

\text{Altitudes, right bisectors and medians are each concurrent}

6.4Area of a triangular regionOften examined

The area of a plane figure is the amount of surface it covers. When the three vertices of a triangle are known as coordinates, the area can be found without measuring any side or height: it is half of a 3\times3 determinant built from the coordinates.

Derivation Area of a triangle from its vertices

Aim. Find the area of \triangle ABC with A(x_1,y_1), B(x_2,y_2), C(x_3,y_3).

Perpendiculars BD, AE, CF to the x-axis make three trapeziums: ABDE, AEFC and BCFD.

Draw perpendiculars BD, AE and CF from the vertices to the x-axis, so D(x_2,0), E(x_1,0), F(x_3,0). This forms three trapeziums ABDE, AEFC and BCFD, each with two vertical parallel sides.
The two upper trapeziums together cover the triangle and also the region under BC. That region is the trapezium BCFD, so subtract it.
\triangle ABC=\text{Trap }ABDE+\text{Trap }AEFC-\text{Trap }BCFD\qquad(1)
Area of a trapezium =\tfrac12\times(sum of the parallel sides)\times(distance between them). The parallel sides are the y-coordinates, and the distances are differences of x-coordinates.
\begin{aligned}\text{Trap }ABDE&=\tfrac12(y_2+y_1)(x_1-x_2)\\ \text{Trap }AEFC&=\tfrac12(y_1+y_3)(x_3-x_1)\\ \text{Trap }BCFD&=\tfrac12(y_2+y_3)(x_3-x_2)\end{aligned}
Substitute in (1) and expand the brackets.
\triangle=\tfrac12\left[x_1y_2-x_2y_2+x_1y_1-x_2y_1+x_3y_1-x_1y_1+x_3y_3-x_1y_3-x_3y_2+x_2y_2-x_3y_3+x_2y_3\right]
The terms x_1y_1, x_2y_2, x_3y_3 cancel. Group the rest by x_1, x_2, x_3.
\triangle=\tfrac12\left[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right]
This bracket is exactly the expansion of a 3\times3 determinant along its third column (a column of 1s).
\triangle=\dfrac12\begin{vmatrix} x_1&y_1&1\\ x_2&y_2&1\\ x_3&y_3&1\end{vmatrix}

Result\text{Area of }\triangle ABC=\dfrac12\left|\,\begin{vmatrix} x_1&y_1&1\\ x_2&y_2&1\\ x_3&y_3&1\end{vmatrix}\,\right|

The determinant is positive when A, B, C are taken anticlockwise and negative when they are taken clockwise. An area cannot be negative, so take the absolute value. If the determinant is zero the 'triangle' is flat: the three points are collinear.

A,\ B,\ C\ \text{collinear}\iff\begin{vmatrix} x_1&y_1&1\\ x_2&y_2&1\\ x_3&y_3&1\end{vmatrix}=0

Use this to prove three points collinear, or to find an unknown coordinate that makes them collinear.

Example 4

Find the area of the triangle whose vertices are (4,-2), (-2,4) and (5,5).

Take x_1=4,\ y_1=-2, x_2=-2,\ y_2=4, x_3=5,\ y_3=5.
\triangle=\dfrac12\begin{vmatrix} 4&-2&1\\ -2&4&1\\ 5&5&1\end{vmatrix}
Expand along the first row.
=\dfrac12\left[4(4-5)+2(-2-5)+1(-10-20)\right]=\dfrac12\left[-4-14-30\right]=\dfrac12(-48)=-24
The negative sign only means the vertices were taken clockwise. Ignore it.

Area =24 square units.

\text{Area}=24\ \text{square units}

Check point

Find the area of the triangle whose vertices are (1,-1), (-4,6) and (-3,-5).

Use the grouped form of the formula.
\triangle=\tfrac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|
Substitute.
=\tfrac12\left|1(6+5)+(-4)(-5+1)+(-3)(-1-6)\right|=\tfrac12\left|11+16+21\right|=\tfrac12(48)=24

Area =24 square units.

\text{Area}=24\ \text{square units}

Area of a quadrilateral or any polygon

Split the figure into triangles by a diagonal and add their areas. Equivalently, list the vertices in order around the boundary, (x_1,y_1),(x_2,y_2),\dots,(x_n,y_n), and use the 'stair' (shoelace) rule:

\text{Area}=\dfrac12\left|(x_1y_2-x_2y_1)+(x_2y_3-x_3y_2)+\dots+(x_ny_1-x_1y_n)\right|

Challenge: area of a parallelogram

Find the area of the parallelogram with vertices E(-1,5), F(2,8), G(4,4) and H(1,1).

The diagonal EG splits the parallelogram into two congruent triangles, EFG and EGH. Find the area of \triangle EFG.
\triangle EFG=\tfrac12\left|(-1)(8-4)+2(4-5)+4(5-8)\right|=\tfrac12\left|-4-2-12\right|=9
Double it.
\text{Area}=2\times9=18
Check with the stair rule, taking E,F,G,H in order.
\tfrac12\left|(-8-10)+(8-32)+(4-4)+(5+1)\right|=\tfrac12\left|-36\right|=18

The diagonal EG cuts the parallelogram into two triangles of equal area.

\text{Area}=18\ \text{square units}

Area of a triangle bounded by three lines

Find the three vertices first: each vertex is the intersection of two of the lines. Then apply the area formula. Ex 6.1 Q14, Review Q9 and Q10 use this method.

6.5Homogeneous equations in two variables

Let f(x,y)=0 be an equation in two variables. Replace x by kx and y by ky. If the whole expression is simply multiplied by k^n, the equation is called **homogeneous of degree n**.

f(kx,ky)=k^{n}f(x,y)\qquad(1)

k is any real number and n is a positive integer, the degree.

Homogeneous linear equation (6.5.1)

An equation of the form ax+by=0 is a homogeneous linear equation in two variables. It is homogeneous of degree n=1, and its line always passes through the origin, because x=0, y=0 satisfies it.

f(kx,ky)=a(kx)+b(ky)=k(ax+by)=k\,f(x,y)\qquad(4)

Check point

Show that 2x+3y=0 is a homogeneous linear equation in two variables.

Let f(x,y)=2x+3y. Replace x by kx and y by ky.
f(kx,ky)=2kx+3ky=k(2x+3y)=k^{1}f(x,y)
This has the form f(kx,ky)=k^nf(x,y) with n=1, so the equation is homogeneous of degree 1, that is, a homogeneous linear equation.

f(kx,ky)=k\,f(x,y)

Non-homogeneous linear equation (6.5.2)

An equation of the form ax+by+c=0 with c\ne0 is a non-homogeneous linear equation, for example 2x-5y+7=0. Here f(kx,ky)=k(2x-5y)+7, which is not k\,f(x,y) because the constant is not multiplied by k. Its line does not pass through the origin.

A homogeneous linear equation always passes through the origin; a non-homogeneous one does not.

Recognising homogeneous equations
Equation	Homogeneous?	Reason
x-4y=0	yes, degree 1	both terms have degree 1
2x-5y+7=0	no	constant term 7
6x^2-4xy+8y^2=0	yes, degree 2	every term has degree 2
x^2+5x=0	no	degrees 2 and 1 are mixed
xy^2+y^3=0	yes, degree 3	every term has degree 3
5x-5=0	no	constant term

6.5.3Joint equation of a pair of lines

Take two lines L_1:\ a_1x+b_1y+c_1=0 (5) and L_2:\ a_2x+b_2y+c_2=0 (6). A product is zero exactly when one of its factors is zero. So the single equation L_1L_2=0 is satisfied by every point of L_1 and by every point of L_2, and by no other point. It represents both lines together and is called their joint equation.

(a_1x+b_1y+c_1)(a_2x+b_2y+c_2)=0\qquad(7)

Multiplying out (7) gives an equation of the second degree, usually written as

ax^2+2hxy+by^2+2gx+2fy+c=0\qquad(8)

The factors 2 in front of h, g, f make later formulas tidy. Equation (8) can be resolved back into the two linear equations.

If both lines pass through the origin (c_1=c_2=0), the lines are a_1x+b_1y=0 (9) and a_2x+b_2y=0 (10), a homogeneous linear system. Their joint equation is

(a_1x+b_1y)(a_2x+b_2y)=a_1a_2x^2+(a_1b_2+a_2b_1)xy+b_1b_2y^2=0\qquad(11)

Put a_1a_2=a, a_1b_2+a_2b_1=2h and b_1b_2=b. This gives the general second-degree homogeneous equation, where a, h, b are not all zero:

ax^2+2hxy+by^2=0\qquad(12)

Any point (x,y) that satisfies a_1x+b_1y=0 or a_2x+b_2y=0 also satisfies (12). For example, 6x^2-4xy+8y^2=0 is a homogeneous quadratic equation in x and y.

Derivation Sum and product of the slopes

Aim. Express the sum and product of the slopes of the lines ax^2+2hxy+by^2=0 in terms of a, h, b (with b\ne0).

Write the two lines through the origin as y=m_1x and y=m_2x, where m_1, m_2 are their slopes. Their joint equation is the product.
(y-m_1x)(y-m_2x)=0\ \Longrightarrow\ y^2-(m_1+m_2)xy+m_1m_2x^2=0\qquad(13)
Divide (12) by b so that the coefficient of y^2 is 1, like (13).
y^2+\dfrac{2h}{b}xy+\dfrac{a}{b}x^2=0
Both equations describe the same pair of lines, so the matching coefficients are equal.
-(m_1+m_2)=\dfrac{2h}{b},\qquad m_1m_2=\dfrac{a}{b}

Resultm_1+m_2=-\dfrac{2h}{b},\qquad m_1m_2=\dfrac{a}{b}

These two results give the angle between the lines (6.5.5) without finding the lines themselves. The book writes the comparison as \dfrac{m_1m_2}{a}=\dfrac{-(m_1+m_2)}{2h}=\dfrac1b.

Joint equation of two given lines

Find the joint equation of the lines x-2y=0 and 3x+y=0, and check the sum and product of their slopes.

Multiply the two left sides.
(x-2y)(3x+y)=3x^2+xy-6xy-2y^2=3x^2-5xy-2y^2
So the joint equation is 3x^2-5xy-2y^2=0, with a=3, 2h=-5, b=-2.
The slopes are \tfrac12 and -3. Compare with the formulas.
m_1+m_2=\tfrac12-3=-\tfrac52=-\dfrac{2h}{b}=-\dfrac{-5}{-2},\qquad m_1m_2=-\tfrac32=\dfrac ab=\dfrac{3}{-2}

The single equation 3x^2-5xy-2y^2=0 is satisfied by every point of both lines.

3x^2-5xy-2y^2=0

6.5.4Homogeneous quadratic equations: a pair of lines through the originOften examined

Taking n=2 in (1) gives a homogeneous quadratic equation: f(kx,ky)=k^2f(x,y) (14). The general one is f(x,y)=ax^2+2hxy+by^2=0 (15).

f(kx,ky)=a(kx)^2+2h(kx)(ky)+b(ky)^2=k^2(ax^2+2hxy+by^2)=k^2f(x,y)\qquad(16)

Derivation Theorem 6.4

Aim. Prove that every homogeneous second-degree equation ax^2+2hxy+by^2=0 (1) represents a pair of lines through the origin, which are real and distinct if h^2>ab, real and coincident if h^2=ab, and imaginary if h^2<ab.

Assumptions.

a\ne0. If a=0 the equation is y(2hx+by)=0, which already shows the two real lines y=0 and 2hx+by=0.

Take a\ne0 and multiply (1) by a, so that the x^2 term becomes a perfect square a^2x^2.
a^2x^2+2ahxy+aby^2=0
Complete the square: add and subtract h^2y^2.
a^2x^2+2ahxy+h^2y^2-h^2y^2+aby^2=0
The first three terms are (ax+hy)^2; take -y^2 common from the last two.
(ax+hy)^2-y^2(h^2-ab)=0
Write the second term as a square.
(ax+hy)^2-\left(y\sqrt{h^2-ab}\right)^2=0
Use p^2-q^2=(p+q)(p-q).
\left(ax+hy+y\sqrt{h^2-ab}\right)\left(ax+hy-y\sqrt{h^2-ab}\right)=0\qquad(2)
Each factor is a homogeneous linear expression, so (1) is the joint equation of two lines through the origin.
ax+\left(h+\sqrt{h^2-ab}\right)y=0\ (3),\qquad ax+\left(h-\sqrt{h^2-ab}\right)y=0\ (4)
The nature of the lines depends on the number h^2-ab under the square root. If it is positive, (3) and (4) are two different real lines. If it is zero, both become ax+hy=0: the same line counted twice (coincident). If it is negative, the square root is imaginary, so the lines are imaginary.

Resultax^2+2hxy+by^2=0\iff\left[ax+\left(h+\sqrt{h^2-ab}\right)y\right]\left[ax+\left(h-\sqrt{h^2-ab}\right)y\right]=0

The sign of h^2-ab plays the same role as the discriminant of a quadratic equation: it decides whether the two lines are real and different, the same, or imaginary.

Nature of the lines ax^2+2hxy+by^2=0
h^2-ab	Lines	Example
>0	real and distinct	x^2-7xy+12y^2=0: \tfrac{49}{4}-12=\tfrac14>0
=0	real and coincident	4x^2-12xy+9y^2=0: 36-36=0, it is (2x-3y)^2=0
<0	imaginary	x^2+xy+y^2=0: \tfrac14-1<0

How to find the two lines

Split the middle term (factorise), as in Example 5. This is the quickest method when the factors have whole-number coefficients.
Solve for the slope. Divide by x^2 and put m=\dfrac yx: bm^2+2hm+a=0. Its two roots are the slopes, and the lines are y=m_1x and y=m_2x.
Use the formula ax+\left(h\pm\sqrt{h^2-ab}\right)y=0 from Theorem 6.4. This always works, including irrational cases.

Example 5

Find the straight lines represented by x^2-7xy+12y^2=0.

Split -7xy into -3xy-4xy (because (-3)(-4)=12, the product of the end coefficients 1\times12).
x^2-3xy-4xy+12y^2=0
Take common factors in pairs.
x(x-3y)-4y(x-3y)=0\ \Longrightarrow\ (x-3y)(x-4y)=0
Each factor gives a line through the origin.
x-3y=0\qquad\text{or}\qquad x-4y=0

x^2-7xy+12y^2=0 is the pair of lines y=\tfrac13x and y=\tfrac14x through the origin.

x-3y=0,\qquad x-4y=0

Lines with irrational slopes

Find the lines represented by x^2+2xy-y^2=0.

Here a=1, 2h=2 so h=1, and b=-1. The number under the root is positive, so the lines are real and distinct.
h^2-ab=1-(1)(-1)=2
Use the formula ax+\left(h\pm\sqrt{h^2-ab}\right)y=0.
x+\left(1+\sqrt2\right)y=0,\qquad x+\left(1-\sqrt2\right)y=0
Their slopes are -\dfrac{1}{1+\sqrt2}=1-\sqrt2 and -\dfrac{1}{1-\sqrt2}=1+\sqrt2. Their product is 1-2=-1, so the lines are perpendicular; this matches a+b=0.

The lines of x^2+2xy-y^2=0 have irrational slopes 1-\sqrt2 and 1+\sqrt2. They are perpendicular because a+b=0.

x+\left(1\pm\sqrt2\right)y=0

6.5.5Angle between the lines ax^2+2hxy+by^2=0Often examined

Theorem 6.4 showed that ax^2+2hxy+by^2=0 represents two lines through the origin. Their angle can be found from a, h, b alone, using the sum and product of the slopes.

Derivation Angle between the pair of lines

Aim. Find the acute angle \theta between the lines represented by ax^2+2hxy+by^2=0.

Both lines pass through the origin; \theta is the acute angle between them.

Let the slopes be m_1 and m_2. From the joint equation (6.5.3):
m_1+m_2=-\dfrac{2h}{b},\qquad m_1m_2=\dfrac{a}{b}
The acute angle between two lines with slopes m_1, m_2 is given by
\tan\theta=\left|\dfrac{m_1-m_2}{1+m_1m_2}\right|
Find m_1-m_2 from the sum and product, using (m_1-m_2)^2=(m_1+m_2)^2-4m_1m_2.
(m_1-m_2)^2=\dfrac{4h^2}{b^2}-\dfrac{4a}{b}=\dfrac{4(h^2-ab)}{b^2}\ \Longrightarrow\ |m_1-m_2|=\dfrac{2\sqrt{h^2-ab}}{|b|}
Also write the denominator with a common denominator.
1+m_1m_2=1+\dfrac ab=\dfrac{a+b}{b}
Divide. The b cancels.
\tan\theta=\dfrac{2\sqrt{h^2-ab}/|b|}{|a+b|/|b|}=\dfrac{2\sqrt{h^2-ab}}{|a+b|}\qquad(3)

Result\tan\theta=\dfrac{2\sqrt{h^2-ab}}{|a+b|}

The formula needs only the coefficients. It is valid even when b=0 (one of the lines is then the y-axis), which can be checked by finding the lines directly.

Special cases.

(i) \theta=0: then \tan\theta=0, so h^2-ab=0, that is h^2=ab. This is the condition for the lines to be coincident.
(ii) \theta=90^\circ: then \tan\theta has no finite value, so the denominator is zero: a+b=0. This is the condition for the lines to be orthogonal (perpendicular): the sum of the coefficients of x^2 and y^2 is zero. It also follows from m_1m_2=\dfrac ab=-1.

Example 6

Find the measure of the acute angle between the lines represented by 6x^2-xy-y^2=0.

Compare with ax^2+2hxy+by^2=0.
a=6,\qquad 2h=-1\Rightarrow h=-\tfrac12,\qquad b=-1
Substitute in \tan\theta=\dfrac{2\sqrt{h^2-ab}}{|a+b|}.
\tan\theta=\dfrac{2\sqrt{\frac14+6}}{|6-1|}=\dfrac{2\cdot\frac52}{5}=1
So \theta=\tan^{-1}(1).
\theta=45^\circ
Check by factorising: 6x^2-xy-y^2=(3x+y)(2x-y), slopes -3 and 2, and \left|\dfrac{2-(-3)}{1+(2)(-3)}\right|=1.

6x^2-xy-y^2=0 is the pair y=2x and y=-3x. The acute angle between them is 45^\circ.

\theta=45^\circ

Derivation Lines through the origin perpendicular to a given pair

Aim. Find the joint equation of the lines through the origin perpendicular to the lines ax^2+2hxy+by^2=0.

Write the given pair as (a_1x+b_1y)(a_2x+b_2y)=0, so that a_1a_2=a, a_1b_2+a_2b_1=2h, b_1b_2=b.
The line through the origin perpendicular to a_1x+b_1y=0 (slope -\tfrac{a_1}{b_1}) is b_1x-a_1y=0 (slope \tfrac{b_1}{a_1}), because the product of the slopes is -1. Similarly for the second line.
b_1x-a_1y=0,\qquad b_2x-a_2y=0
Multiply to get their joint equation.
(b_1x-a_1y)(b_2x-a_2y)=b_1b_2x^2-(a_1b_2+a_2b_1)xy+a_1a_2y^2=0
Replace the products by a, h, b.
bx^2-2hxy+ay^2=0

Result\text{Perpendicular pair of } ax^2+2hxy+by^2=0:\quad bx^2-2hxy+ay^2=0

Swap the coefficients of x^2 and y^2 and change the sign of the xy term. The book finds these lines by factorising first (Example 7); this shortcut gives the same answer in one line and is a good check.

Example 7

Find a joint equation of the straight lines through the origin and perpendicular to the lines represented by 3x^2+5xy+2y^2=0.

Factorise the given equation by splitting 5xy=3xy+2xy.
3x^2+3xy+2xy+2y^2=3x(x+y)+2y(x+y)=(x+y)(3x+2y)=0
So the given lines are x+y=0, that is y=-x (1), and 3x+2y=0, that is y=-\tfrac32x (2).
A line through the origin perpendicular to y=mx is y=-\dfrac1mx. Perpendicular to (1):
y=x\quad\text{or}\quad y-x=0\qquad(3)
Perpendicular to (2):
y=\tfrac23x\quad\text{or}\quad 3y-2x=0\qquad(4)
Multiply (3) and (4) to get the joint equation.
(y-x)(3y-2x)=3y^2-2xy-3xy+2x^2=0\ \Longrightarrow\ 2x^2-5xy+3y^2=0
Check with the shortcut bx^2-2hxy+ay^2=0 with a=3, 2h=5, b=2: it gives 2x^2-5xy+3y^2=0.

Solid: the given lines 3x^2+5xy+2y^2=0. Dashed: the perpendicular lines 2x^2-5xy+3y^2=0 (same colour means perpendicular partners).

2x^2-5xy+3y^2=0

6.6Applications of analytic geometry

Analytic geometry is used in physics and engineering, and also in aviation, rocketry, space science and spaceflight. It is the foundation of most modern fields of geometry, including algebraic, differential, discrete and computational geometry. The three examples below use the three main tools of this unit: concurrency, area and the angle between a pair of lines.

Example 8: three roads

An engineer needs to design a traffic intersection where three straight roads meet. The roads are 2x-4y+5=0, 7x-8y+7=0 and 4x+4y-11=0. Determine whether the roads are concurrent and, if so, find the point of intersection of the roads.

Label the roads 2x-4y+5=0 (1), 7x-8y+7=0 (2), 4x+4y-11=0 (3). Multiply (1) by -2 so that the y terms of (1) and (2) cancel.
-4x+8y-10=0\qquad(4)
Add (4) and (2).
3x-3=0\ \Longrightarrow\ x=1
Substitute in (1).
2(1)-4y+5=0\ \Longrightarrow\ -4y=-7\ \Longrightarrow\ y=\tfrac74
So roads (1) and (2) meet at \left(1,\tfrac74\right). Substitute this point in (3).
4(1)+4\left(\tfrac74\right)-11=4+7-11=0
The third road also passes through \left(1,\tfrac74\right), so the three roads are concurrent there.

The three roads meet at \left(1,\tfrac74\right).

\text{Concurrent at }\left(1,\ \tfrac74\right)

Example 9: area of a park

A triangular park is bounded by three straight roads that meet at the points A(2,3), B(8,5) and C(5,10). A city planner needs the area of the park for landscaping. How can he determine it?

Use the determinant formula with x_1=2,\ y_1=3,\ x_2=8,\ y_2=5,\ x_3=5,\ y_3=10.
\triangle=\dfrac12\begin{vmatrix} 2&3&1\\ 8&5&1\\ 5&10&1\end{vmatrix}
Expand along the first row.
=\dfrac12\left[2(5-10)-3(8-5)+1(80-25)\right]=\dfrac12\left[-10-9+55\right]=\dfrac12(36)=18

The triangular park has area 18 square units.

\text{Area of the park}=18\ \text{square units}

Example 10: railway tracks

A railway engineer is designing an intersection where two rail tracks meet at a junction. The joint equation of the tracks is x^2-4xy-12y^2=0. Determine the angle between the individual lines representing the tracks.

Factorise by splitting -4xy=2xy-6xy (because 2\times(-6)=-12).
x^2+2xy-6xy-12y^2=x(x+2y)-6y(x+2y)=(x+2y)(x-6y)=0
The tracks are x+2y=0 and x-6y=0, that is y=-\tfrac12x and y=\tfrac16x.
m_1=-\tfrac12,\qquad m_2=\tfrac16
Use the angle formula.
\tan\theta=\left|\dfrac{m_2-m_1}{1+m_1m_2}\right|=\dfrac{\frac16+\frac12}{1-\frac{1}{12}}=\dfrac{\frac23}{\frac{11}{12}}=\dfrac{8}{11}=0.727
Take the inverse tangent.
\theta=\tan^{-1}(0.727)=36.03^\circ
Check with \tan\theta=\dfrac{2\sqrt{h^2-ab}}{|a+b|}, where a=1, h=-2, b=-12.
\tan\theta=\dfrac{2\sqrt{4+12}}{|1-12|}=\dfrac{8}{11}

The two tracks cross at the origin at an acute angle of about 36^\circ.

\theta\approx36.03^\circ

Formula sheet

Distance formula

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

d: distance between (x_1,y_1) and (x_2,y_2)

Mid-point formula

M=\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right)

M: mid-point of the segment joining (x_1,y_1) and (x_2,y_2)

Section formula (internal division)

P=\left(\dfrac{m_1x_2+m_2x_1}{m_1+m_2},\ \dfrac{m_1y_2+m_2y_1}{m_1+m_2}\right)

P divides AB so that AP:PB=m_1:m_2; A(x_1,y_1), B(x_2,y_2)

Slope of a line

m=\tan\alpha=\dfrac{y_2-y_1}{x_2-x_1}

\alpha: inclination; (x_1,y_1), (x_2,y_2): two points of the line

Parallel and perpendicular lines (slopes)

l_1\parallel l_2\iff m_1=m_2,\qquad l_1\perp l_2\iff m_1m_2=-1

m_1, m_2: slopes of the lines

Slope-intercept and point-slope forms

y=mx+c,\qquad y-y_1=m(x-x_1)

m: slope, c: y-intercept, (x_1,y_1): a point of the line

Two-point and two-intercept forms

\dfrac{y-y_1}{y_2-y_1}=\dfrac{x-x_1}{x_2-x_1},\qquad \dfrac xa+\dfrac yb=1

(x_1,y_1), (x_2,y_2): two points; a, b: x- and y-intercepts

General form of a line

ax+by+c=0:\quad m=-\dfrac ab,\quad x\text{-int}=-\dfrac ca,\quad y\text{-int}=-\dfrac cb

a, b not both zero

Angle between two lines

\tan\theta=\left|\dfrac{m_2-m_1}{1+m_1m_2}\right|

\theta: acute angle between lines of slopes m_1 and m_2

Distance of a point from a line

d=\dfrac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}

d: perpendicular distance of (x_1,y_1) from ax+by+c=0

Position of a point relative to a line

b>0:\quad ax_1+by_1+c\ \begin{cases}>0 & \text{above}\\ =0 & \text{on}\\ <0 & \text{below}\end{cases}

(x_1,y_1): the point; line ax+by+c=0 written with b>0

Point of intersection of two lines

\left(\dfrac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1},\ \dfrac{a_2c_1-a_1c_2}{a_1b_2-a_2b_1}\right)

lines a_1x+b_1y+c_1=0 and a_2x+b_2y+c_2=0 with a_1b_2-a_2b_1\ne0

Parallel and perpendicular lines (coefficients)

l_1\parallel l_2\iff a_1b_2=a_2b_1,\qquad l_1\perp l_2\iff a_1a_2+b_1b_2=0

a_i, b_i: coefficients of x and y in the two lines

Condition of concurrency of three lines

\begin{vmatrix} a_1&b_1&c_1\\ a_2&b_2&c_2\\ a_3&b_3&c_3\end{vmatrix}=0

lines a_ix+b_iy+c_i=0, i=1,2,3, not all parallel

Centroid of a triangle

G=\left(\dfrac{x_1+x_2+x_3}{3},\ \dfrac{y_1+y_2+y_3}{3}\right)

(x_i,y_i): the vertices; G divides each median in the ratio 2:1 from the vertex

Altitude from A (general form)

(x_3-x_2)x+(y_3-y_2)y-x_1(x_3-x_2)-y_1(y_3-y_2)=0

altitude through A(x_1,y_1) perpendicular to BC, with B(x_2,y_2), C(x_3,y_3)

Right bisector of BC (general form)

(x_3-x_2)x+(y_3-y_2)y-\tfrac12(x_3^2-x_2^2)-\tfrac12(y_3^2-y_2^2)=0

B(x_2,y_2), C(x_3,y_3)

Area of a triangle

\triangle=\dfrac12\left|\,\begin{vmatrix} x_1&y_1&1\\ x_2&y_2&1\\ x_3&y_3&1\end{vmatrix}\,\right|=\dfrac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|

(x_i,y_i): the vertices

square units

Collinear points

\begin{vmatrix} x_1&y_1&1\\ x_2&y_2&1\\ x_3&y_3&1\end{vmatrix}=0

the three points lie on one line exactly when the area is zero

Area of a polygon (stair rule)

\text{Area}=\dfrac12\left|\sum_{i=1}^{n}(x_iy_{i+1}-x_{i+1}y_i)\right|,\quad (x_{n+1},y_{n+1})=(x_1,y_1)

vertices taken in order around the boundary

square units

Homogeneous equation of degree n

f(kx,ky)=k^nf(x,y)

k: any real number; n: positive integer (degree)

Joint equation of two lines through the origin

(a_1x+b_1y)(a_2x+b_2y)=0\iff ax^2+2hxy+by^2=0

a=a_1a_2, 2h=a_1b_2+a_2b_1, b=b_1b_2

Sum and product of slopes

m_1+m_2=-\dfrac{2h}{b},\qquad m_1m_2=\dfrac ab

m_1, m_2: slopes of the lines ax^2+2hxy+by^2=0, b\ne0

Pair of lines from the general second-degree equation

\begin{vmatrix} a&h&g\\ h&b&f\\ g&f&c\end{vmatrix}=0

condition for ax^2+2hxy+by^2+2gx+2fy+c=0 to represent two straight lines

Lines represented by ax^2+2hxy+by^2=0

ax+\left(h\pm\sqrt{h^2-ab}\right)y=0

a\ne0

Nature of the lines

h^2>ab:\ \text{real, distinct};\quad h^2=ab:\ \text{coincident};\quad h^2<ab:\ \text{imaginary}

for ax^2+2hxy+by^2=0

Angle between the lines ax^2+2hxy+by^2=0

\tan\theta=\dfrac{2\sqrt{h^2-ab}}{|a+b|}

\theta: acute angle between the two lines

Perpendicular (orthogonal) pair

a+b=0

the lines ax^2+2hxy+by^2=0 are perpendicular exactly when the coefficients of x^2 and y^2 add to zero

Pair perpendicular to a given pair

bx^2-2hxy+ay^2=0

lines through the origin perpendicular to the lines ax^2+2hxy+by^2=0

Key terms

Analytic (coordinate) geometry: The branch of mathematics that studies geometric shapes using coordinates and equations, combining algebra and geometry.
Slope: The rise per unit run of a non-vertical line, m=\tan\alpha=\dfrac{y_2-y_1}{x_2-x_1}.
Inclination: The angle \alpha (0^\circ\le\alpha<180^\circ) that a line makes with the positive x-axis, measured anticlockwise.
Collinear points: Points that lie on one straight line; for three points the area determinant is zero.
Concurrent lines: Three or more lines that pass through one common point.
Point of concurrency: The common point where three or more concurrent lines intersect.
Median: The segment joining a vertex of a triangle to the mid-point of the opposite side.
Altitude: The perpendicular from a vertex of a triangle to the opposite side (extended if necessary).
Right (perpendicular) bisector: The line through the mid-point of a side of a triangle, perpendicular to that side.
Angle bisector: The line through a vertex that divides the angle at that vertex into two equal angles.
Centroid: The point of intersection of the three medians; G=\left(\tfrac{x_1+x_2+x_3}{3},\tfrac{y_1+y_2+y_3}{3}\right), dividing each median in the ratio 2:1 from the vertex.
Circumcentre: The point of intersection of the three right bisectors; it is equidistant from the three vertices and is the centre of the circumcircle.
Orthocentre: The point of intersection of the three altitudes of a triangle.
Incentre: The point of intersection of the three angle bisectors; it is equidistant from the three sides.
Euler line: The line through the orthocentre, centroid and circumcentre of a triangle; the centroid divides the orthocentre-circumcentre segment in the ratio 2:1.
Homogeneous equation of degree n: An equation f(x,y)=0 with f(kx,ky)=k^nf(x,y) for every real k; every term has the same total degree n and there is no constant term.
Homogeneous linear equation: An equation ax+by=0; its line always passes through the origin.
Non-homogeneous linear equation: An equation ax+by+c=0 with c\ne0; its line does not pass through the origin.
Joint equation: A single equation, the product of the equations of two lines, that represents both lines together.
Second-degree homogeneous equation: An equation ax^2+2hxy+by^2=0 (a, h, b not all zero); it represents a pair of lines through the origin.
Coincident lines: Two lines that are the same line; for ax^2+2hxy+by^2=0 this happens when h^2=ab.
Imaginary lines: Lines of ax^2+2hxy+by^2=0 when h^2<ab; the only real point on them is the origin.
Orthogonal lines: Perpendicular lines; the pair ax^2+2hxy+by^2=0 is orthogonal when a+b=0.

Exercise 6.1p. 171–172

Concurrency by the determinant or the direct method; equations of altitudes (negative reciprocal slope through a vertex), right bisectors (negative reciprocal slope through a mid-point) and medians (vertex to mid-point); area and collinearity from the area determinant.

Which sets of lines are concurrent in the given figure? Also, tell the point of concurrency.

Solution

(i)

Figure (i): lines p, q, l, m, n with points A to E.

Concurrent lines are three or more lines through one point, so look at each labelled point and count the lines passing through it.
At A: the lines p, q and m pass through A. Three lines, so they are concurrent at A.
At B: the lines p, n and l pass through B. They are concurrent at B.
At C only q and l meet, at D only q and n meet, and at E only m and n meet. Two lines always meet (unless parallel), so these points are just points of intersection, not points of concurrency.

p,\ q,\ m \text{ concurrent at } A;\qquad p,\ n,\ l \text{ concurrent at } B

(ii)

Figure (ii): lines a, b, c, m, p, q crossing the line r at A, C and D.

At A: the lines a, b, c and r all pass through A. Four lines, so they are concurrent at A.
At D: the lines p, q and r pass through D. They are concurrent at D.
At C only the two lines m and r meet, so C is not a point of concurrency.

a,\ b,\ c,\ r \text{ concurrent at } A;\qquad p,\ q,\ r \text{ concurrent at } D

Check whether the lines are concurrent or not.

Solution

(i) 3x-4y-13=0,\quad 8x-11y-33=0,\quad 2x-3y-7=0

Write the determinant of the coefficients a, b, c of the three lines and expand along the first row.
\begin{aligned}D&=\begin{vmatrix} 3 & -4 & -13\\ 8 & -11 & -33\\ 2 & -3 & -7\end{vmatrix}\\ &=3\left[(-11)(-7)-(-33)(-3)\right]-(-4)\left[(8)(-7)-(-33)(2)\right]+(-13)\left[(8)(-3)-(-11)(2)\right]\\ &=3(-22)-(-4)(10)+(-13)(-2)\\ &=-66+40+26=0\end{aligned}
The determinant is zero, and the first two lines have different slopes, so the three lines are concurrent.
Find the common point from the first two lines by the cross-multiplication rule (or by elimination).
x=\dfrac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}=11,\qquad y=\dfrac{a_2c_1-a_1c_2}{a_1b_2-a_2b_1}=5
Check in the third line: 2(11)-3(5)-7=0.

The three lines pass through (11,5).

\text{Concurrent at } (11,5)

(ii) x+2y-4=0,\quad x-y-1=0,\quad 4x+5y-13=0

Write the determinant of the coefficients a, b, c of the three lines and expand along the first row.
\begin{aligned}D&=\begin{vmatrix} 1 & 2 & -4\\ 1 & -1 & -1\\ 4 & 5 & -13\end{vmatrix}\\ &=1\left[(-1)(-13)-(-1)(5)\right]-2\left[(1)(-13)-(-1)(4)\right]+(-4)\left[(1)(5)-(-1)(4)\right]\\ &=1(18)-2(-9)+(-4)(9)\\ &=18+18-36=0\end{aligned}
The determinant is zero, and the first two lines have different slopes, so the three lines are concurrent.
Find the common point from the first two lines by the cross-multiplication rule (or by elimination).
x=\dfrac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}=2,\qquad y=\dfrac{a_2c_1-a_1c_2}{a_1b_2-a_2b_1}=1
Check in the third line: 4(2)+5(1)-13=0.

The three lines pass through (2,1).

\text{Concurrent at } (2,1)

Determine the value of a if the lines 2x-y+3=0, x-y=0, 3x+ay+1=0 are concurrent.

Solution

For concurrent lines the determinant of coefficients is zero. The second line x-y=0 has c=0.
\begin{vmatrix} 2&-1&3\\ 1&-1&0\\ 3&a&1\end{vmatrix}=0
Expand along the first row.
2\left[(-1)(1)-(0)(a)\right]-(-1)\left[(1)(1)-(0)(3)\right]+3\left[(1)(a)-(-1)(3)\right]=0
Simplify.
2(-1)+1(1)+3(a+3)=0\ \Longrightarrow\ -2+1+3a+9=0\ \Longrightarrow\ 3a+8=0
Solve for a.
a=-\dfrac{8}{3}
Check: x-y=0 gives y=x; then 2x-x+3=0 gives x=-3, so the first two lines meet at (-3,-3). In the third line: 3(-3)-\tfrac83(-3)+1=-9+8+1=0.

With a=-\tfrac83 all three lines pass through (-3,-3).

a=-\dfrac{8}{3}

Show that the lines x+y+1=0, x-y+1=0 and x-axis are concurrent. Also prove that x-axis bisects the angle between x+y+1=0 and x-y+1=0.

Solution

Concurrency

The equation of the x-axis is y=0, that is 0x+1y+0=0. Write the determinant of coefficients.
\begin{aligned}D&=\begin{vmatrix} 1 & 1 & 1\\ 1 & -1 & 1\\ 0 & 1 & 0\end{vmatrix}\\ &=1\left[(-1)(0)-(1)(1)\right]-1\left[(1)(0)-(1)(0)\right]+1\left[(1)(1)-(-1)(0)\right]\\ &=1(-1)-1(0)+1(1)\\ &=-1+0+1=0\end{aligned}
The determinant is zero (and the lines are not parallel), so the three lines are concurrent.
The common point: put y=0 in x+y+1=0 to get x=-1. The point (-1,0) also satisfies x-y+1=0, since -1-0+1=0.
\text{Point of concurrency}=(-1,0)

The x-axis bisects the angle

Slopes: x-y+1=0 gives y=x+1, slope 1, inclination 45^\circ. And x+y+1=0 gives y=-x-1, slope -1, inclination 135^\circ.
The product of the slopes is (1)(-1)=-1, so the two lines are perpendicular: the angle between them is 90^\circ.
At (-1,0) the line x-y+1=0 rises at 45^\circ above the x-axis, and the line x+y+1=0 falls at 45^\circ below it (because 180^\circ-135^\circ=45^\circ).
So the x-axis lies inside the right angle between them and makes 45^\circ with each line. It divides the angle into two equal parts, so it bisects it.
\angle(\text{line}_1,\ x\text{-axis})=\angle(x\text{-axis},\ \text{line}_2)=45^\circ
Another way to see it: replacing y by -y changes x+y+1=0 into x-y+1=0. The two lines are mirror images of each other in the x-axis, so the x-axis is the bisector of the angle between them.

The lines meet on the x-axis at (-1,0) and make equal angles of 45^\circ with it.

\text{Concurrent at }(-1,0);\ \ x\text{-axis bisects the right angle between the lines}

Find the equations of altitudes and their point of concurrency in the triangle ABC when A(4,-2), B(5,5) and C(-1,3). What is the name of point of concurrency?

Solution

**Altitude AD (from A, perpendicular to BC).** Slope of BC, then the negative reciprocal.
m_{BC}=\dfrac{3-5}{-1-5}=\dfrac{-2}{-6}=\dfrac13\quad\Longrightarrow\quad m_{AD}=-3
Point-slope form through A(4,-2).
y+2=-3(x-4)\ \Longrightarrow\ y+2=-3x+12\ \Longrightarrow\ 3x+y-10=0\qquad(1)
**Altitude BE (from B, perpendicular to CA).**
m_{CA}=\dfrac{-2-3}{4-(-1)}=\dfrac{-5}{5}=-1\quad\Longrightarrow\quad m_{BE}=1
Through B(5,5).
y-5=1(x-5)\ \Longrightarrow\ x-y=0\qquad(2)
**Altitude CF (from C, perpendicular to AB).**
m_{AB}=\dfrac{5-(-2)}{5-4}=7\quad\Longrightarrow\quad m_{CF}=-\dfrac17
Through C(-1,3); multiply by 7 to clear the fraction.
y-3=-\dfrac17(x+1)\ \Longrightarrow\ 7y-21=-x-1\ \Longrightarrow\ x+7y-20=0\qquad(3)
Point of concurrency. From (2), y=x. Put it in (1).
3x+x-10=0\ \Longrightarrow\ x=\tfrac52,\qquad y=\tfrac52
Check in (3): \tfrac52+7\left(\tfrac52\right)-20=\tfrac{5+35-40}{2}=0. So all three altitudes pass through \left(\tfrac52,\tfrac52\right).
The point where the altitudes meet is called the orthocentre.

The three altitudes meet at the orthocentre H\left(\tfrac52,\tfrac52\right).

3x+y-10=0,\ \ x-y=0,\ \ x+7y-20=0;\quad \text{orthocentre }\left(\tfrac52,\tfrac52\right)

Find the equations of right bisectors and their point of concurrency in the triangle XYZ when X(0,0), Y(7,0) and Z(7,4). What is the name of point of concurrency?

Solution

**Right bisector of XY.** Mid-point of XY is \left(\tfrac72,0\right). XY lies on the x-axis (horizontal), so its right bisector is the vertical line through the mid-point.
x=\tfrac72\ \Longrightarrow\ 2x-7=0\qquad(1)
**Right bisector of YZ.** Mid-point of YZ is (7,2). YZ is vertical (x=7), so its right bisector is horizontal.
y=2\ \Longrightarrow\ y-2=0\qquad(2)
**Right bisector of XZ.** Mid-point \left(\tfrac72,2\right); slope of XZ is \tfrac47, so the bisector has slope -\tfrac74.
y-2=-\dfrac74\left(x-\dfrac72\right)
Multiply by 8 and simplify.
8y-16=-14x+49\ \Longrightarrow\ 14x+8y-65=0\qquad(3)
Point of concurrency. (1) and (2) give x=\tfrac72, y=2. Check in (3): 14\left(\tfrac72\right)+8(2)-65=49+16-65=0.
The point where the right bisectors meet is the circumcentre \left(\tfrac72,2\right). The angle at Y is a right angle, so, as expected, the circumcentre is the mid-point of the hypotenuse XZ. The circumradius is \sqrt{\left(\tfrac72\right)^2+2^2}=\tfrac{\sqrt{65}}{2}.

The right bisectors meet at the circumcentre, the mid-point of the hypotenuse XZ.

2x-7=0,\ \ y-2=0,\ \ 14x+8y-65=0;\quad \text{circumcentre }\left(\tfrac72,2\right)

Find the equations of medians and their point of concurrency in the triangle DEF when D(-6,-4), E(6,-4) and F(-2,4). What is the name of point of concurrency?

Solution

**Median from D.** Mid-point of EF: \left(\tfrac{6-2}{2},\tfrac{-4+4}{2}\right)=(2,0). Two-point form through D(-6,-4) and (2,0).
\dfrac{y+4}{0+4}=\dfrac{x+6}{2+6}\ \Longrightarrow\ 8(y+4)=4(x+6)\ \Longrightarrow\ x-2y-2=0\qquad(1)
**Median from E.** Mid-point of DF: \left(\tfrac{-6-2}{2},\tfrac{-4+4}{2}\right)=(-4,0). Through E(6,-4) and (-4,0).
\dfrac{y+4}{0+4}=\dfrac{x-6}{-4-6}\ \Longrightarrow\ -10(y+4)=4(x-6)\ \Longrightarrow\ 2x+5y+8=0\qquad(2)
**Median from F.** Mid-point of DE: \left(\tfrac{-6+6}{2},\tfrac{-4-4}{2}\right)=(0,-4). Through F(-2,4) and (0,-4).
\dfrac{y-4}{-4-4}=\dfrac{x+2}{0+2}\ \Longrightarrow\ 2(y-4)=-8(x+2)\ \Longrightarrow\ 4x+y+4=0\qquad(3)
Point of concurrency. From (1), x=2y+2. Put it in (2).
\begin{aligned}2(2y+2)+5y+8&=0\\ \Longrightarrow\ 9y+12&=0\\ \Longrightarrow\ y&=-\tfrac43\\ x&=-\tfrac83+2=-\tfrac23\end{aligned}
Check in (3): 4\left(-\tfrac23\right)-\tfrac43+4=-\tfrac{12}{3}+4=0. Also by the formula \left(\tfrac{-6+6-2}{3},\tfrac{-4-4+4}{3}\right)=\left(-\tfrac23,-\tfrac43\right).
The point where the medians meet is called the centroid.

The medians meet at the centroid G\left(-\tfrac23,-\tfrac43\right).

x-2y-2=0,\ \ 2x+5y+8=0,\ \ 4x+y+4=0;\quad \text{centroid }\left(-\tfrac23,-\tfrac43\right)

Prove that (a) altitudes, (b) right bisectors and (c) medians of the following triangles are concurrent.

Solution

(i) A(4,6), B(7,2), C(2,3)

Slopes of the sides: m_{BC}=\dfrac{3-2}{2-7}=-\dfrac15, m_{CA}=\dfrac{6-3}{4-2}=\dfrac32, m_{AB}=\dfrac{2-6}{7-4}=-\dfrac43. Mid-points: BC\left(\tfrac92,\tfrac52\right), CA\left(3,\tfrac92\right), AB\left(\tfrac{11}2,4\right).

(a) Altitudes

Altitude from A: slope 5 (perpendicular to BC), through A(4,6).
y-6=5(x-4)\ \Longrightarrow\ 5x-y-14=0\qquad(1)
Altitude from B: slope -\tfrac23, through B(7,2).
3(y-2)=-2(x-7)\ \Longrightarrow\ 2x+3y-20=0\qquad(2)
Altitude from C: slope \tfrac34, through C(2,3).
4(y-3)=3(x-2)\ \Longrightarrow\ 3x-4y+6=0\qquad(3)
Determinant of the coefficients of (1), (2), (3), expanded along the first row.
\begin{aligned}D&=\begin{vmatrix} 5 & -1 & -14\\ 2 & 3 & -20\\ 3 & -4 & 6\end{vmatrix}\\ &=5\left[(3)(6)-(-20)(-4)\right]-(-1)\left[(2)(6)-(-20)(3)\right]+(-14)\left[(2)(-4)-(3)(3)\right]\\ &=5(-62)-(-1)(72)+(-14)(-17)\\ &=-310+72+238=0\end{aligned}
The determinant is zero, so the three lines are concurrent. Solving two of them gives the orthocentre H\left(\tfrac{62}{17},\tfrac{72}{17}\right).

The orthocentre is H\left(\tfrac{62}{17},\tfrac{72}{17}\right).

(b) Right bisectors

Right bisector of BC: slope 5, through \left(\tfrac92,\tfrac52\right).
y-\tfrac52=5\left(x-\tfrac92\right)\ \Longrightarrow\ 5x-y-20=0\qquad(1)
Right bisector of CA: slope -\tfrac23, through \left(3,\tfrac92\right); multiply by 6.
6y-27=-4x+12\ \Longrightarrow\ 4x+6y-39=0\qquad(2)
Right bisector of AB: slope \tfrac34, through \left(\tfrac{11}2,4\right); multiply by 8.
8y-32=6x-33\ \Longrightarrow\ 6x-8y-1=0\qquad(3)
Determinant of the coefficients of (1), (2), (3), expanded along the first row.
\begin{aligned}D&=\begin{vmatrix} 5 & -1 & -20\\ 4 & 6 & -39\\ 6 & -8 & -1\end{vmatrix}\\ &=5\left[(6)(-1)-(-39)(-8)\right]-(-1)\left[(4)(-1)-(-39)(6)\right]+(-20)\left[(4)(-8)-(6)(6)\right]\\ &=5(-318)-(-1)(230)+(-20)(-68)\\ &=-1590+230+1360=0\end{aligned}
The determinant is zero, so the three lines are concurrent. Solving two of them gives the circumcentre O\left(\tfrac{159}{34},\tfrac{115}{34}\right).

The circumcentre is O\left(\tfrac{159}{34},\tfrac{115}{34}\right).

(c) Medians

Median from A(4,6) to \left(\tfrac92,\tfrac52\right): slope \dfrac{5/2-6}{9/2-4}=-7.
y-6=-7(x-4)\ \Longrightarrow\ 7x+y-34=0\qquad(1)
Median from B(7,2) to \left(3,\tfrac92\right): slope \dfrac{9/2-2}{3-7}=-\tfrac58.
8(y-2)=-5(x-7)\ \Longrightarrow\ 5x+8y-51=0\qquad(2)
Median from C(2,3) to \left(\tfrac{11}2,4\right): slope \dfrac{4-3}{11/2-2}=\tfrac27.
7(y-3)=2(x-2)\ \Longrightarrow\ 2x-7y+17=0\qquad(3)
Determinant of the coefficients of (1), (2), (3), expanded along the first row.
\begin{aligned}D&=\begin{vmatrix} 7 & 1 & -34\\ 5 & 8 & -51\\ 2 & -7 & 17\end{vmatrix}\\ &=7\left[(8)(17)-(-51)(-7)\right]-1\left[(5)(17)-(-51)(2)\right]+(-34)\left[(5)(-7)-(8)(2)\right]\\ &=7(-221)-1(187)+(-34)(-51)\\ &=-1547-187+1734=0\end{aligned}
The determinant is zero, so the three lines are concurrent. Solving two of them gives the centroid G\left(\tfrac{13}{3},\tfrac{11}{3}\right).

The centroid is G\left(\tfrac{13}{3},\tfrac{11}{3}\right).

H\left(\tfrac{62}{17},\tfrac{72}{17}\right),\quad O\left(\tfrac{159}{34},\tfrac{115}{34}\right),\quad G\left(\tfrac{13}{3},\tfrac{11}{3}\right)

(ii) P(-4,0), Q(2,0), R(0,3)

Slopes: m_{PQ}=0 (PQ is on the x-axis), m_{QR}=\dfrac{3-0}{0-2}=-\dfrac32, m_{RP}=\dfrac{0-3}{-4-0}=\dfrac34. Mid-points: PQ(-1,0), QR\left(1,\tfrac32\right), RP\left(-2,\tfrac32\right).

(a) Altitudes

Altitude from P: slope \tfrac23 (perpendicular to QR), through P(-4,0).
3y=2(x+4)\ \Longrightarrow\ 2x-3y+8=0\qquad(1)
Altitude from Q: slope -\tfrac43 (perpendicular to RP), through Q(2,0).
3y=-4(x-2)\ \Longrightarrow\ 4x+3y-8=0\qquad(2)
Altitude from R: PQ is horizontal, so this altitude is the vertical line through R(0,3).
x=0\qquad(3)
Determinant of the coefficients of (1), (2), (3), expanded along the first row.
\begin{aligned}D&=\begin{vmatrix} 2 & -3 & 8\\ 4 & 3 & -8\\ 1 & 0 & 0\end{vmatrix}\\ &=2\left[(3)(0)-(-8)(0)\right]-(-3)\left[(4)(0)-(-8)(1)\right]+8\left[(4)(0)-(3)(1)\right]\\ &=2(0)-(-3)(8)+8(-3)\\ &=0+24-24=0\end{aligned}
The determinant is zero, so the three lines are concurrent. Solving two of them gives the orthocentre H\left(0,\tfrac83\right).

The orthocentre is H\left(0,\tfrac83\right).

(b) Right bisectors

Right bisector of QR: slope \tfrac23, through \left(1,\tfrac32\right); multiply by 6.
6y-9=4x-4\ \Longrightarrow\ 4x-6y+5=0\qquad(1)
Right bisector of RP: slope -\tfrac43, through \left(-2,\tfrac32\right); multiply by 6.
6y-9=-8x-16\ \Longrightarrow\ 8x+6y+7=0\qquad(2)
Right bisector of PQ: vertical line through (-1,0).
x+1=0\qquad(3)
Determinant of the coefficients of (1), (2), (3), expanded along the first row.
\begin{aligned}D&=\begin{vmatrix} 4 & -6 & 5\\ 8 & 6 & 7\\ 1 & 0 & 1\end{vmatrix}\\ &=4\left[(6)(1)-(7)(0)\right]-(-6)\left[(8)(1)-(7)(1)\right]+5\left[(8)(0)-(6)(1)\right]\\ &=4(6)-(-6)(1)+5(-6)\\ &=24+6-30=0\end{aligned}
The determinant is zero, so the three lines are concurrent. Solving two of them gives the circumcentre O\left(-1,\tfrac16\right).

The circumcentre is O\left(-1,\tfrac16\right).

(c) Medians

Median from P(-4,0) to \left(1,\tfrac32\right): slope \dfrac{3/2}{5}=\tfrac{3}{10}.
10y=3(x+4)\ \Longrightarrow\ 3x-10y+12=0\qquad(1)
Median from Q(2,0) to \left(-2,\tfrac32\right): slope \dfrac{3/2}{-4}=-\tfrac38.
8y=-3(x-2)\ \Longrightarrow\ 3x+8y-6=0\qquad(2)
Median from R(0,3) to (-1,0): slope \dfrac{0-3}{-1-0}=3.
y-3=3x\ \Longrightarrow\ 3x-y+3=0\qquad(3)
Determinant of the coefficients of (1), (2), (3), expanded along the first row.
\begin{aligned}D&=\begin{vmatrix} 3 & -10 & 12\\ 3 & 8 & -6\\ 3 & -1 & 3\end{vmatrix}\\ &=3\left[(8)(3)-(-6)(-1)\right]-(-10)\left[(3)(3)-(-6)(3)\right]+12\left[(3)(-1)-(8)(3)\right]\\ &=3(18)-(-10)(27)+12(-27)\\ &=54+270-324=0\end{aligned}
The determinant is zero, so the three lines are concurrent. Solving two of them gives the centroid G\left(-\tfrac23,1\right).

The centroid is G\left(-\tfrac23,1\right).

H\left(0,\tfrac83\right),\quad O\left(-1,\tfrac16\right),\quad G\left(-\tfrac23,1\right)

Find the area of triangles.

Solution

(i) A(1,1), B(4,5), C(12,-1)

Use \triangle=\tfrac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right| with the vertices in the given order.
\triangle=\tfrac12\left|1\left(5-(-1)\right)+4\left(-1-1\right)+12\left(1-5\right)\right|
Multiply out and add.
\triangle=\tfrac12\left|6-8-48\right|=\tfrac12(50)=25

Area =25 square units.

\text{Area}=25\ \text{square units}

(ii) D(3,1), E(2,3), F(2,2)

Use \triangle=\tfrac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right| with the vertices in the given order.
\triangle=\tfrac12\left|3\left(3-2\right)+2\left(2-1\right)+2\left(1-3\right)\right|
Multiply out and add.
\triangle=\tfrac12\left|3+2-4\right|=\tfrac12(1)=\tfrac{1}{2}

Area =\tfrac{1}{2} square units.

\text{Area}=\tfrac{1}{2}\ \text{square units}

By finding area of triangle, show that points A(6,0), B(-3,6) and C(3,2) are collinear.

Solution

Find the area of the 'triangle' ABC with the determinant formula.
\triangle=\dfrac12\begin{vmatrix} 6&0&1\\ -3&6&1\\ 3&2&1\end{vmatrix}
Expand along the first row.
=\dfrac12\left[6(6-2)-0(-3-3)+1(-6-18)\right]=\dfrac12\left[24-0-24\right]=0
The area is zero, so no triangle is formed: the three points lie on one straight line. They are collinear.
Check: the line through A and B has slope \dfrac{6-0}{-3-6}=-\dfrac23, so it is 2x+3y-12=0, and C gives 2(3)+3(2)-12=0.

All three points lie on the line 2x+3y-12=0.

\triangle=0\ \Longrightarrow\ A,\ B,\ C\ \text{are collinear}

Using the formula of area, find x if the points P(3,2), Q(-1,x), R(7,3) are collinear.

Solution

Collinear points give zero area, so the determinant is zero.
\begin{vmatrix} 3&2&1\\ -1&x&1\\ 7&3&1\end{vmatrix}=0
Expand along the first row.
3(x-3)-2(-1-7)+1(-3-7x)=0
Simplify.
3x-9+16-3-7x=0\ \Longrightarrow\ -4x+4=0\ \Longrightarrow\ x=1
Check: slope PR=\dfrac{3-2}{7-3}=\dfrac14 and slope PQ=\dfrac{1-2}{-1-3}=\dfrac14. The slopes are equal, so the points are collinear.

With x=1 the points lie on the line x-4y+5=0.

x=1

Vertices of a triangle are (3,-4), (4,h) and (2,6). Find h if area of the triangle is 10 square units.

Solution

Area =\tfrac12|D|, so the determinant must be +20 or -20 (the sign depends on the order of the vertices).
\tfrac12\left|\begin{vmatrix} 3&-4&1\\ 4&h&1\\ 2&6&1\end{vmatrix}\right|=10
Expand along the first row.
D=3(h-6)+4(4-2)+1(24-2h)=3h-18+8+24-2h=h+14
Solve |h+14|=20.
h+14=20\ \Longrightarrow\ h=6\qquad\text{or}\qquad h+14=-20\ \Longrightarrow\ h=-34
Both values give area 10, so both are answers. Check h=6: \tfrac12|6+14|=10. Check h=-34: \tfrac12|-34+14|=10.

h=6: vertices (3,-4), (4,6), (2,6), area 10.

h=-34: a long thin triangle, also of area 10 (axes not to the same scale).

h=6\quad\text{or}\quad h=-34

Note. The question does not fix the order of the vertices, so both signs of the determinant must be allowed; that is why there are two answers.

Find the area of following figures.

Solution

(i)

Figure (i)

Read the vertices from the figure: (-3,-1), (2,-1) and (2,3).
Apply the area formula.
\triangle=\tfrac12\left|(-3)(-1-3)+2(3-(-1))+2(-1-(-1))\right|=\tfrac12\left|12+8+0\right|=10
Check: the triangle has a right angle at (2,-1), with base 5 (from x=-3 to x=2) and height 4 (from y=-1 to y=3), so the area is \tfrac12(5)(4)=10.

10\ \text{square units}

(ii)

Figure (ii)

Read the vertices: (1,3), (-2,-3) and (4,-3).
Apply the area formula.
\triangle=\tfrac12\left|1(-3-(-3))+(-2)(-3-3)+4(3-(-3))\right|=\tfrac12\left|0+12+24\right|=18
Check: base from (-2,-3) to (4,-3) is 6; the height from y=-3 up to y=3 is 6; area =\tfrac12(6)(6)=18.

18\ \text{square units}

Find the area of triangle bounded by the lines:

Solution

(i) 4x-5y+7=0, x-2=0 and y+1=0

Vertices. Each vertex is where two of the lines meet. x=2 and y=-1 meet at (2,-1).
4x-5y+7=0 with x=2: 8-5y+7=0, so y=3. Vertex (2,3).
4x-5y+7=0 with y=-1: 4x+5+7=0, so x=-3. Vertex (-3,-1).
Area.
\triangle=\tfrac12\left|2(3-(-1))+2(-1-(-1))+(-3)(-1-3)\right|=\tfrac12\left|8+0+12\right|=10
This is the same right-angled triangle as Q13 (i): base 5, height 4.

The three lines enclose a right-angled triangle of area 10.

10\ \text{square units}

(ii) x-2y-6=0, 3x-y+3=0 and 2x+y-4=0

Vertex 1: x-2y-6=0 and 3x-y+3=0. From the first, x=2y+6; substitute.
3(2y+6)-y+3=0\ \Longrightarrow\ 5y=-21\ \Longrightarrow\ y=-\tfrac{21}{5},\quad x=-\tfrac{12}{5}
Vertex 2: x-2y-6=0 and 2x+y-4=0.
2(2y+6)+y-4=0\ \Longrightarrow\ 5y=-8\ \Longrightarrow\ y=-\tfrac85,\quad x=\tfrac{14}{5}
Vertex 3: 3x-y+3=0 and 2x+y-4=0. Add them.
5x-1=0\ \Longrightarrow\ x=\tfrac15,\quad y=3x+3=\tfrac{18}{5}
Area. Every coordinate has denominator 5, so take \tfrac15 out of each coordinate: this puts \tfrac1{25} in front of the determinant.
\triangle=\dfrac12\cdot\dfrac1{25}\left|(-12)(-8-18)+14(18+21)+1(-21+8)\right|
Evaluate.
=\dfrac1{50}\left|312+546-13\right|=\dfrac{845}{50}=\dfrac{169}{10}=16.9

The three lines enclose a triangle of area 16.9 square units.

\dfrac{169}{10}=16.9\ \text{square units}

Exercise 6.2p. 176–177

Every question uses ax^2+2hxy+by^2=0: factorise (or use y=mx) to get the lines, h^2-ab to decide their nature, \tan\theta=\dfrac{2\sqrt{h^2-ab}}{|a+b|} for the angle, and a+b=0 for perpendicular lines. Remember that h is half the coefficient of xy.

Find the lines represented by each of the following joint equations.

Solution

(i) x^2+5xy+6y^2=0

The equation is homogeneous of degree 2 (every term has degree 2), so it stands for two lines through the origin. Factorise the left side by splitting the middle term.
x^2+2xy+3xy+6y^2=x(x+2y)+3y(x+2y)=(x+2y)(x+3y)
A product is zero only when one of its factors is zero. Each factor, put equal to zero, is one line.
(x+2y)(x+3y)=0\ \Longrightarrow\ x+2y=0\ \text{ or }\ x+3y=0
So the joint equation represents the two lines
x+2y=0\quad\text{and}\quad x+3y=0

Both lines pass through the origin, with slopes -\tfrac12 and -\tfrac13.

x+2y=0\quad\text{and}\quad x+3y=0

(ii) 7x^2-2xy-9y^2=0

The equation is homogeneous of degree 2 (every term has degree 2), so it stands for two lines through the origin. Factorise the left side by splitting the middle term.
7x^2+7xy-9xy-9y^2=7x(x+y)-9y(x+y)=(x+y)(7x-9y)
A product is zero only when one of its factors is zero. Each factor, put equal to zero, is one line.
(x+y)(7x-9y)=0\ \Longrightarrow\ x+y=0\ \text{ or }\ 7x-9y=0
So the joint equation represents the two lines
x+y=0\quad\text{and}\quad 7x-9y=0

Slopes -1 and \tfrac79.

x+y=0\quad\text{and}\quad 7x-9y=0

(iii) x^2+6xy=0

There is no y^2 term (b=0), but the equation is still homogeneous of degree 2. Take out the common factor x.
x^2+6xy=x(x+6y)=0
Each factor equal to zero gives one line.
x=0\ \text{ or }\ x+6y=0
x=0 is the y-axis (a vertical line, which has no slope). This is why the slope method y=mx misses one line when b=0: always factorise in this case.

One line is the y-axis; the other has slope -\tfrac16.

x=0\ (\text{the } y\text{-axis})\quad\text{and}\quad x+6y=0

(iv) x^2-8xy+12y^2=0

The equation is homogeneous of degree 2 (every term has degree 2), so it stands for two lines through the origin. Factorise the left side by splitting the middle term.
x^2-2xy-6xy+12y^2=x(x-2y)-6y(x-2y)=(x-2y)(x-6y)
A product is zero only when one of its factors is zero. Each factor, put equal to zero, is one line.
(x-2y)(x-6y)=0\ \Longrightarrow\ x-2y=0\ \text{ or }\ x-6y=0
So the joint equation represents the two lines
x-2y=0\quad\text{and}\quad x-6y=0

Slopes \tfrac12 and \tfrac16.

x-2y=0\quad\text{and}\quad x-6y=0

(v) 5x^2+3xy-2y^2=0

The equation is homogeneous of degree 2 (every term has degree 2), so it stands for two lines through the origin. Factorise the left side by splitting the middle term.
5x^2+5xy-2xy-2y^2=5x(x+y)-2y(x+y)=(x+y)(5x-2y)
A product is zero only when one of its factors is zero. Each factor, put equal to zero, is one line.
(x+y)(5x-2y)=0\ \Longrightarrow\ x+y=0\ \text{ or }\ 5x-2y=0
So the joint equation represents the two lines
x+y=0\quad\text{and}\quad 5x-2y=0

Slopes -1 and \tfrac52.

x+y=0\quad\text{and}\quad 5x-2y=0

(vi) x^2-3xy-y^2=0

No two whole numbers have product 1\times(-1)=-1 and sum -3, so the middle term cannot be split. Put y=mx (every line through the origin except x=0 has this form) and divide by x^2.
x^2-3x(mx)-(mx)^2=0\ \Longrightarrow\ 1-3m-m^2=0\ \Longrightarrow\ m^2+3m-1=0
Solve the quadratic in m by the quadratic formula.
m=\dfrac{-3\pm\sqrt{9+4}}{2}=\dfrac{-3\pm\sqrt{13}}{2}
Each value of m gives one line y=mx. Multiply by 2 to clear fractions.
2y=\left(-3+\sqrt{13}\right)x\quad\text{and}\quad 2y=\left(-3-\sqrt{13}\right)x
Check with the formula ax+\left(h\pm\sqrt{h^2-ab}\right)y=0 where a=1, h=-\tfrac32, b=-1: h^2-ab=\tfrac94+1=\tfrac{13}4, so the lines are x+\left(\tfrac{-3\pm\sqrt{13}}{2}\right)y=0, which is the same pair written the other way round (the slopes \tfrac{-2}{-3\pm\sqrt{13}} simplify to \tfrac{-3\mp\sqrt{13}}{2}).

Slopes about 0.30 and -3.30; their product is -1 because a+b=1-1=0, so these lines are perpendicular.

\left(3-\sqrt{13}\right)x+2y=0\quad\text{and}\quad \left(3+\sqrt{13}\right)x+2y=0

Find measure of acute angle between lines represented by the following equations.

Solution

(i) x^2-y^2=0

Compare with ax^2+2hxy+by^2=0.
a=1,\qquad h=0,\qquad b=-1
Here a+b=1+(-1)=0, so the denominator of \tan\theta=\dfrac{2\sqrt{h^2-ab}}{|a+b|} is zero: \tan\theta is not defined.
\tan\theta=\dfrac{2\sqrt{0+1}}{0}\ \ \text{not defined}\ \Longrightarrow\ \theta=90^\circ
Check: x^2-y^2=(x-y)(x+y)=0 gives y=x and y=-x, slopes 1 and -1, and (1)(-1)=-1.

The lines y=x and y=-x are perpendicular.

\theta=90^\circ

(ii) x^2+5xy+4y^2=0

Compare with ax^2+2hxy+by^2=0. The coefficient of xy is 2h, so halve it to get h.
a=1,\qquad h=\tfrac52,\qquad b=4
Work out h^2-ab (it must be positive for real, distinct lines).
h^2-ab=\tfrac{25}{4}-4=\tfrac94,\qquad \sqrt{h^2-ab}=\tfrac32
Use \tan\theta=\dfrac{2\sqrt{h^2-ab}}{|a+b|} for the acute angle \theta.
\tan\theta=\dfrac{2\left(\tfrac32\right)}{|1+4|}=\dfrac35
Take the inverse tangent.
\theta=\tan^{-1}\left(\tfrac35\right)\ \Longrightarrow\ \theta\approx30.96^\circ
Check by factorising: x^2+5xy+4y^2=(x+y)(x+4y), slopes m_1=-1, m_2=-\tfrac14, and \left|\dfrac{m_2-m_1}{1+m_1m_2}\right|=\left|\dfrac{3/4}{5/4}\right|=\tfrac35.