CHAPTER ONE: COORDINATE GEOMETRY

Form 4 Mathematics • Comprehensive Notes

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  2. Topic 2Areas and PerimeterIn the app
  3. Topic 3Three Dimensional FiguresIn the app
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Chapter Objectives

By the end of this chapter, you should be able to:

  1. Derive the general equation of a straight line.
  2. Determine the coordinates of the mid-point of a line segment.
  3. Calculate the distance between two points on the xy-plane.
  4. Determine the conditions for both parallel and perpendicular lines.
  5. Solve problems involving parallel and perpendicular lines.
  6. Solve problems on parallel and perpendicular lines in real-life situations.

Key Terms and Concepts

Coordinate Geometry (Analytic Geometry): The study of geometry using a coordinate system. It links algebra and geometry by describing geometric shapes with algebraic equations.
Gradient (Slope) $m$: A measure of the steepness of a line. It is the ratio of the change in $y$ to the change in $x$: $m = \dfrac{\Delta y}{\Delta x} = \dfrac{y_2 - y_1}{x_2 - x_1}$.
y-intercept ($c$): The point where a line crosses the y-axis (when $x = 0$).
x-intercept: The point where a line crosses the x-axis (when $y = 0$).
Equation of a Straight Line: The algebraic relationship satisfied by all points on the line. Common forms: $y = mx + c$ (slope-intercept) and $ax + by + c = 0$ (standard form).
Mid-point: The point that divides a line segment into two equal parts. Formula: $M\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$.
Distance Formula: The length of the line segment joining two points: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
Parallel Lines: Lines that never meet and have the same gradient ($m_1 = m_2$).
Perpendicular Lines: Lines that intersect at right angles ($90^\circ$). Their gradients satisfy $m_1 m_2 = -1$.

1. Equation of a Straight Line

Consider two points $(2, 1)$ and $(2, 3)$ as shown below.

Figure 1.1 - Line through (2,1) and (2,3)
Figure 1.1

The gradient $m$ of the line through these points is:

$$ m = \frac{\text{change in } y}{\text{change in } x} = \frac{3-1}{2-(-2)} = \frac{2}{4} = \frac{1}{2} $$

If $(x, y)$ is another point on the line, the gradient is also $\dfrac{y-3}{x-2}$. Since the gradient is constant:

$$ \frac{y-3}{x-2} = \frac{1}{2} \implies y-3 = \frac{1}{2}(x-2) \implies y = \frac{1}{2}x + 2 $$

In general, if a line has gradient $m$ and y-intercept $c$, its equation is:

$$ y = mx + c $$

General Derivation (Two-Point Form)

For two known points $(x_1, y_1)$ and $(x_2, y_2)$ and an unknown point $(x, y)$ on the line:

$$ m = \frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1} $$

Cross-multiplying and rearranging leads to the standard form:

$$ ax + by + c = 0 $$
Figure 1.2 - Two points on a line
Figure 1.2

SPOTLIGHT

Did you know? The slope-intercept form $y = mx + c$ is one of the most used equations in real life — from predicting stock prices and modelling population growth to designing roads and ski slopes! The gradient $m$ tells engineers how steep a ramp or roof should be.

Worked Examples

Example 1.1

Find the equation of a straight line which passes through $(2, 3)$ and whose slope is $\dfrac{2}{3}$.

Solution:

Using $m = \dfrac{y - y_1}{x - x_1}$:

$$ \frac{2}{3} = \frac{y-3}{x-2} \implies y-3 = \frac{2}{3}(x-2) \implies y = \frac{2}{3}x + \frac{5}{3} $$

Alternatively using $y = mx + c$: $3 = \dfrac{2}{3}(2) + c \implies c = \dfrac{5}{3}$.

∴ Equation is $y = \dfrac{2}{3}x + \dfrac{5}{3}$.

Example 1.2

Find the equation of a straight line passing through the points $(0, -1)$ and $(2, 4)$ in the form $ax + by + c = 0$.

Solution:

$$ m = \frac{4 - (-1)}{2 - 0} = \frac{5}{2} $$ $$ \frac{y + 1}{x} = \frac{5}{2} \implies 5x = 2y + 2 \implies 5x - 2y - 2 = 0 $$
Example 1.3

Find the equation of a straight line passing through $(5, 8)$ and having the same y-intercept as the line $2x - y = 5$. Write it in standard form.

Solution:

From $2x - y = 5$, when $x=0$: $y = -5$. So y-intercept is $-5$.

Line passes through $(0, -5)$ and $(5, 8)$:

$$ m = \frac{8 - (-5)}{5 - 0} = \frac{13}{5} $$ $$ \frac{y + 5}{x} = \frac{13}{5} \implies 13x = 5y + 25 \implies 13x - 5y - 25 = 0 $$
Example 1.4

Find the equation of a straight line with gradient $\dfrac{3}{2}$ and y-intercept $5$ in the form $ax + by + c = 0$.

Solution:

$$ y = \frac{3}{2}x + 5 \implies 2y = 3x + 10 \implies 3x - 2y + 10 = 0 $$
Example 1.5

Find the equation of a straight line with x-intercept $2$ and y-intercept $-5$.

Solution: Passes through $(2, 0)$ and $(0, -5)$.

$$ \frac{y + 5}{x} = \frac{0 - (-5)}{2 - 0} = \frac{5}{2} \implies y = \frac{5}{2}x - 5 $$

Jog your Brain – Equation of a Straight Line

Test yourself! Click to reveal answers after attempting.

Q1. Find the equation of a straight line joining the points $(3, 6)$ and $(2, 4)$ in the form $y = mx + c$.

$m = \dfrac{4-6}{2-3} = \dfrac{-2}{-1} = 2$. Using point $(3,6)$: $6 = 2(3) + c \implies c = 0$. Equation: $y = 2x$.

Q2. Find the equation of a straight line passing through $(0, 2)$ with gradient $2$.

$y = 2x + 2$.

Q3. Calculate the equation of a straight line passing through $(2, -4)$ which has the same slope as the line $3x - 6y + 12 = 0$.

From $3x - 6y + 12 = 0$: $6y = 3x + 12 \implies y = \dfrac{1}{2}x + 2$, so $m = \dfrac{1}{2}$. Then $-4 = \dfrac{1}{2}(2) + c \implies c = -5$. Equation: $y = \dfrac{1}{2}x - 5$.

Q4. Find the equation of a straight line with slope $\dfrac{1}{3}$ having the same y-intercept as $2x - 5y + 10 = 0$.

From $2x - 5y + 10 = 0$: $-5y = -2x - 10 \implies y = \dfrac{2}{5}x + 2$, y-intercept $= 2$. Equation: $y = \dfrac{1}{3}x + 2$.

Q5. Find the equation of a straight line through $(5, 0)$ with infinite gradient.

Infinite gradient means a vertical line: $x = 5$.

Q6. Find the equation of a straight line through $(0, 4)$ with gradient zero.

Horizontal line: $y = 4$.

Mid-point of a Line Segment

The mid-point $M$ of a line segment $AB$ is the point which divides the segment into two equal halves. Thus $\overline{AM} = \overline{MB}$.

Figure 1.3 - Mid-point illustration
Figure 1.3

Using similar triangles (or section formula with ratio $1:1$), the coordinates of the mid-point $M(x, y)$ are:

$$ x = \frac{x_1 + x_2}{2}, \quad y = \frac{y_1 + y_2}{2} $$ $$ \therefore M\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) $$
Figure 1.4 - Similar triangles for mid-point
Figure 1.4 – Derivation using similar triangles

SPOTLIGHT

In a parallelogram, the diagonals bisect each other — their mid-points coincide! This is a powerful way to prove a quadrilateral is a parallelogram using coordinate geometry.

Worked Examples

Example 1.6

Find the mid-point of the line segments whose end points are:

(a) $A(8, -2)$ and $B(-4, 3)$

(b) $A(1, 4)$ and $B(3, 6)$

(c) $A(a + 2b, 2a - b)$ and $B(b - 3a, -3b)$

Solution:

(a) $M = \left( \dfrac{8 + (-4)}{2}, \dfrac{-2 + 3}{2} \right) = (2, \dfrac{1}{2})$

(b) $M = \left( \dfrac{1+3}{2}, \dfrac{4+6}{2} \right) = (2, 5)$

(c) $M = \left( \dfrac{a+2b + b-3a}{2}, \dfrac{2a-b + (-3b)}{2} \right) = \left( \dfrac{-2a + 3b}{2}, a - 2b \right)$ or $(1.5b - a, a - 2b)$

Example 1.7

$M$ is the mid-point of $AB$. If $M(2, 3)$ and $B(-1, 2)$, find the coordinates of $A$.

Solution: Let $A(x_2, y_2)$.

$$ 2 = \frac{-1 + x_2}{2} \implies 4 = -1 + x_2 \implies x_2 = 5 $$ $$ 3 = \frac{2 + y_2}{2} \implies 6 = 2 + y_2 \implies y_2 = 4 $$

∴ $A(5, 4)$.

Example 1.8

A parallelogram has vertices $A(1, 7)$, $B(6, 4)$, $C(3, 1)$ and $D(-2, 4)$.

(a) Find the coordinates of the mid-points of the diagonals.

(b) Show that the diagonals bisect each other.

Solution:

Mid-point of $DB$: $\left( \dfrac{6 + (-2)}{2}, \dfrac{4+4}{2} \right) = (2, 4)$

Mid-point of $AC$: $\left( \dfrac{1+3}{2}, \dfrac{7+1}{2} \right) = (2, 4)$

Since both mid-points are the same, the diagonals bisect each other.

Jog your Brain – Mid-point

Q1. Find the mid-point of $(-3, 4)$ and $(4, -4)$.

$\left( \dfrac{-3+4}{2}, \dfrac{4+(-4)}{2} \right) = \left( \dfrac{1}{2}, 0 \right)$

Q2. $M$ is the mid-point of $AB$. $M(2, 1)$ and $A(-3, 1)$. Find coordinates of $B$.

Let $B(x, y)$. $2 = \dfrac{-3 + x}{2} \implies x = 7$; $1 = \dfrac{1 + y}{2} \implies y = 1$. So $B(7, 1)$.

Q3. Show that the mid-point of $(\sqrt{12}, \sqrt{8})$ and $(4\sqrt{3}, 4\sqrt{2})$ is $(3\sqrt{3}, 3\sqrt{2})$.

$\sqrt{12} = 2\sqrt{3}$, $\sqrt{8} = 2\sqrt{2}$. Mid-point: $\left( \dfrac{2\sqrt{3}+4\sqrt{3}}{2}, \dfrac{2\sqrt{2}+4\sqrt{2}}{2} \right) = (3\sqrt{3}, 3\sqrt{2})$. Proven.

Distance Between Two Points on a Plane

Consider two points $A(x_1, y_1)$ and $B(x_2, y_2)$. Draw a right-angled triangle with hypotenuse $AB$.

Figure 1.5 - Distance derivation
Figure 1.5

By Pythagoras’ Theorem:

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

SPOTLIGHT

The distance formula is simply the Pythagorean theorem in disguise! It is used in GPS systems, computer graphics, robotics path planning, and even in measuring the “distance” between data points in machine learning.

Worked Examples

Example 1.9

Find the distance between $A(-1, -1)$ and $B(2, 3)$.

Solution:

$$ d = \sqrt{(2 - (-1))^2 + (3 - (-1))^2} = \sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5 $$

Distance is $5$ units.

Example 1.10

The distance between $P(k, 2)$ and $Q(-3, 5)$ is $5$. Find the possible values of $k$.

Solution:

$$ 5 = \sqrt{(k + 3)^2 + (2 - 5)^2} $$ $$ 25 = (k+3)^2 + 9 \implies k^2 + 6k + 9 + 9 - 25 = 0 \implies k^2 + 6k - 7 = 0 $$ $$ k = \frac{-6 \pm \sqrt{36 + 28}}{2} = \frac{-6 \pm 8}{2} $$ $$ k = 1 \quad \text{or} \quad k = -7 $$

Jog your Brain – Distance

Q1. Find the distance between $(-2, -3)$ and $(4, 5)$.

$d = \sqrt{(4+2)^2 + (5+3)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$

Q2. Prove that the triangle with vertices $A(1, -1)$, $B(4, -1)$, $C(1, 3)$ is right-angled.

$AB = 3$, $AC = 4$, $BC = 5$. Since $3^2 + 4^2 = 5^2$, it is right-angled at $A$.

Q3. Find the distance of the point $(8, 6)$ from the origin.

$d = \sqrt{8^2 + 6^2} = \sqrt{64+36} = \sqrt{100} = 10$

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2. Parallel Lines

Two or more lines which do not meet (no matter how far extended) and maintain a constant distance between them are called parallel lines. Railway tracks are a classic real-life example.

Figure 1.6 - Parallel lines
Figure 1.6

From similar triangles, it follows that parallel lines have equal slopes:

$$ m_1 = m_2 $$

SPOTLIGHT

In architecture and civil engineering, parallel lines ensure that walls, railway tracks, and power lines remain evenly spaced and structurally sound. Even the lanes on a highway rely on the concept of parallel lines!

Example 1.11

Find the equation of a straight line passing through the origin and parallel to the line $4x - 3y - 2 = 0$. Write it in the form $ax + by + c = 0$.

Solution:

Rewrite: $y = \dfrac{4}{3}x - \dfrac{2}{3}$, so $m_1 = \dfrac{4}{3}$.

Since parallel, $m_2 = \dfrac{4}{3}$. Through $(0,0)$:

$$ \frac{y}{x} = \frac{4}{3} \implies 4x - 3y = 0 $$

Jog your Brain – Parallel Lines

Q1. Find the equation of a straight line passing through $(3, 4)$ and parallel to $12x + 6y = 9$.

$12x + 6y = 9 \implies y = -2x + \dfrac{3}{2}$, $m = -2$. Through $(3,4)$: $4 = -2(3) + c \implies c = 10$. Equation: $y = -2x + 10$ or $2x + y - 10 = 0$.

Q2. Show that $3x - y + 6 = 0$ is parallel to $6x - 2y + 7 = 0$.

First line: $y = 3x + 6$, $m=3$. Second: $2y = 6x + 7 \implies y = 3x + \dfrac{7}{2}$, $m=3$. Same slope ⇒ parallel.

Q3. The line through $A(k, 4)$ and $B(3, 2k)$ is parallel to $y + 3x - 4 = 0$. Find $k$.

Given line: $y = -3x + 4$, $m=-3$. Slope of $AB$: $\dfrac{2k-4}{3-k} = -3 \implies 2k-4 = -3(3-k) \implies 2k-4 = -9 + 3k \implies -4+9 = 3k-2k \implies 5 = k$.

3. Perpendicular Lines

Two lines are perpendicular (normal) if they meet at a right angle ($90^\circ$).

Figure 1.7 - Perpendicular lines
Figure 1.7

Using similar triangles and complementary angles, the condition is:

$$ m_1 m_2 = -1 \quad \text{(or } m_2 = -\dfrac{1}{m_1} \text{)} $$

SPOTLIGHT

Perpendicular lines are everywhere: the corners of a book, the intersection of latitude and longitude lines on a map (approximately), and the axes of a coordinate system itself. In design, right angles create stability and balance.

Example 1.12

Find the equation of a straight line passing through $(-4, 2)$ and perpendicular to $x + 2y = 4$.

Solution:

$x + 2y = 4 \implies y = -\dfrac{1}{2}x + 2$, so $m_1 = -\dfrac{1}{2}$.

Thus $m_2 = 2$. Through $(-4, 2)$:

$$ y - 2 = 2(x + 4) \implies y = 2x + 10 $$
Example 1.13

Find the equation of the perpendicular bisector of the line segment joining $A(9, 1)$ and $B(7, -5)$.

Solution:

Mid-point: $\left( \dfrac{9+7}{2}, \dfrac{1+(-5)}{2} \right) = (8, -2)$

Slope of $AB$: $\dfrac{1-(-5)}{9-7} = 3$. Therefore perpendicular slope $= -\dfrac{1}{3}$.

$$ \frac{y + 2}{x - 8} = -\frac{1}{3} \implies 3(y+2) = -(x-8) \implies x + 3y - 2 = 0 $$

Jog your Brain – Perpendicular Lines

Q1. Find the equation of the perpendicular bisector of the segment joining $(0, 2)$ and $(3, 6)$.

Mid-point: $\left( \dfrac{3}{2}, 4 \right)$. Slope of segment: $\dfrac{6-2}{3-0} = \dfrac{4}{3}$, so perp. slope $= -\dfrac{3}{4}$. Equation: $y-4 = -\dfrac{3}{4}\left(x - \dfrac{3}{2}\right)$. Simplify to $3x + 4y - 22.5 = 0$ or $6x + 8y - 45 = 0$.

Q2. Find the equation of a straight line passing through $(3, -2)$ and perpendicular to the line through $(-4, 3)$ and $(2, 6)$.

Slope of given points: $\dfrac{6-3}{2-(-4)} = \dfrac{3}{6} = \dfrac{1}{2}$. Perp. slope $= -2$. Through $(3,-2)$: $y + 2 = -2(x-3) \implies y = -2x + 4$.

Chapter Summary

Suggested Answers (Selected Exercises)

Exercise 1.1 (Equation of Straight Lines)

1. $y = 2x$
2. $y = 2x + 2$
3. $y = \dfrac{1}{2}x - 5$
4. Slope of joining $(3,3)$ to $(2,4)$ is $-1$. Equation through $(3,3)$: $y = -x + 6$
5. $y = \dfrac{1}{3}x + 2$
6. $x = 5$ (vertical line)
7. $y = 4$ (horizontal line)
9. $y = x + 1$
11. Passes through $(-2,0)$ and $(0,-3)$: $y = -\dfrac{3}{2}x - 3$

Exercise 1.2 / Mid-point Questions

1(a) $\left(\dfrac{1}{2}, 0\right)$    (b) $(5, 2)$    (c) $\left(-\dfrac{1}{2}, -2\right)$
2. $B(7, 1)$

Exercise 1.3 (Distance)

1(a) $10$    (b) $3\sqrt{2}$    (f) $\sqrt{148} = 2\sqrt{37}$
3. $AB=3$, $AC=4$, $BC=5$ → right-angled at $A$
8. $10$ units from origin

Parallel Lines

1. $2x + y - 10 = 0$
2. Both have slope $3$ → parallel
3. $k = 5$

Perpendicular Lines

Example-style: Perpendicular bisector of $(0,2)$ & $(3,6)$: $6x + 8y - 45 = 0$
Through $(3,-2)$ perp. to points $(-4,3),(2,6)$: $y = -2x + 4$

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