7.4.2 Equation of Straight Lines (Part 2) - SPM Additional Mathematics

7.4.2 Equation of Straight Lines

Case 1
1. The gradient and coordinates of a point are given.

2. The equation of a straight line with gradient m passes through the point (x₁, y₁) is:

Example 1:

A straight line with gradient –3 passes through the point (–1, 5). Find the equation of this line.

Solution:

y – y₁= m (x – x₁)

y – 5= – 3 (x – (–1))

y – 5= – 3x – 3

y = – 3x + 2

Case 2
1. The coordinates of two points are given.

2. The equation of a straight line joining the points (x₁, y₁)
and (x₂, y₂) is:

Example 2:
Find the equation of the straight line joining the points (2, 4) and (5, 6).

Solution:

\begin{array}{l} \frac{y - y_{1}}{x - x_{1}} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ Let (x_{1}, y_{1}) = (2, 4) and (x_{2}, y_{2}) = (5, 6) \\ \frac{y - 4}{x - 2} = \frac{6 - 4}{5 - 2} \\ \frac{y - 4}{x - 2} = \frac{2}{3} \\ 3 y - 12 = 2 x - 4 \\ 3 y = 2 x + 8 \end{array}

Case 3

1. The equation of a straight line with x–intercept “a” and y–intercept“b” is:

Example 3:

Find the equation of the straight line joining the points (5, 0) and (0, –6).

Solution:

x–intercept, a = 5, y–intercept, b = –6

Equation of the straight line

\begin{array}{l} \frac{x}{a} + \frac{y}{b} = 1 \\ \frac{x}{5} + \frac{y}{(- 6)} = 1 \\ \frac{x}{5} - \frac{y}{6} = 1 \end{array}

The equation of a straight line can be expressed in three forms:

(a)

(b)

(c)