__6.4 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

*passes through the*

**m** point (

*x*_{1},*y*_{1}) is:**Example 1:**

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

*Solution:**y*–

*y*

_{1 }=

*m*(

*x*–

*x*

_{1})

*y*– 5

_{ }= – 3 (

*x*– (–1))

*y*– 5

_{ }= – 3

*x*– 3

*y*= – 3*x*+ 2

**Case 2****1.**The

**coordinates of two points**are given.

**2.**The equation of a straight line joining the points (

*x*

_{1},

*y*

_{1})

and (

*x*

_{2},

*y*

_{2}) is:

**Example 2:**

Find the equation of the straight line joining the points (2, 4) and

(5, 6).

$$\begin{array}{l}\frac{y-{y}_{1}}{x-{x}_{1}}=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\ \text{Let}\left({x}_{1},{y}_{1}\right)=\left(2,\text{}4\right)\text{and}\left({x}_{2},{y}_{2}\right)\text{=}\left(5,\text{}6\right)\\ \frac{y-4}{x-2}=\frac{6-4}{5-2}\\ \frac{y-4}{x-2}=\frac{2}{3}\\ 3y-12=2x-4\\ 3y=2x+8\end{array}$$

*Solution:*$$\begin{array}{l}\frac{y-{y}_{1}}{x-{x}_{1}}=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\ \text{Let}\left({x}_{1},{y}_{1}\right)=\left(2,\text{}4\right)\text{and}\left({x}_{2},{y}_{2}\right)\text{=}\left(5,\text{}6\right)\\ \frac{y-4}{x-2}=\frac{6-4}{5-2}\\ \frac{y-4}{x-2}=\frac{2}{3}\\ 3y-12=2x-4\\ 3y=2x+8\end{array}$$

**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).

(0, –6).

*Solution:**x–*intercept, a = 5,

*y–*intercept, b = –6

Equation of the straight line

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