6.4 Equation of Straight Lines (Part 2)

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

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

y – y₁ = m (x – x₁)

y – 5 = – 3 (x – (–1))

y – 5 = – 3x – 3

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:
$\begin{array}{l}\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}\\ \text{Let }\left(x_1,y_1\right)=\left(2,4\right)\text{ and }\left(x_2,y_2\right)\text{ = }\left(5,6\right)\\ \frac{y-4}{x-2}=\frac{6-4}{5-2}\\ \frac{y-4}{x-2}=\frac{2}{3}\\ 3y-12=2x-4\\ 3y=2x+8\end{array}$

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

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