Chapter 4, Problem 6PT

To determine

To find:The equation of the line for the given condition.

Answer to Problem 6PT

The equation of the straight line is $y=\frac{-3}{4}x+3$ .

Explanation of Solution

Given:

The equation passes through points $\left(2,5\right)$ and $\left(-2,8\right)$

Calculation:

The standard form of line passing through two points is $y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$ .

Here $\frac{y_2-y_1}{x_2-x_1}$ is the slope of the straight line. $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are the points from which the straight line passes.

Substitute $\left(2,5\right)$ for $\left(x_1,y_1\right)$ and $\left(-2,8\right)$ for $\left(x_2,y_2\right)$ .

The equation of the straight line is

  $\begin{array}{l}y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)\\ y-5=\frac{8-5}{\left(-2\right)-\left(2\right)}\left(x-\left(2\right)\right)\\ y-5=-\frac{3}{4}\left(x-2\right)\\ y=\frac{-3}{4}x-2+5\\ y=\frac{-3}{4}x+3\end{array}$

Therefore, the equation of the straight line is

  $y=\frac{-3}{4}x+3$ .