Chapter 4.2, Problem 28PPS

To determine

To find the equation of line in point-slope form, slope-intercept form and standard form.

Answer to Problem 28PPS

Equation of the line in point-slope form is $y-1=\frac{4}{5}\left(x+5\right)$ .

Equation of the line in slope-intercept form is $y=\frac{4}{5}x+5$ .

Equation of the line in standard form is $4x-5y=-25$

Explanation of Solution

Given:

Line passing through the points $\left(-5,1\right),\left(5,9\right)$ .

Concept used:

Equation of a line passing through the points $\left(x_1,y_1\right)\&\left(x_2,y_2\right)$ is $y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$ .

Calculation:

Given,

  $\left(x_1,y_1\right)=\left(-5,1\right)\&\left(x_2,y_2\right)=\left(5,9\right)$

Now the equation of the line in point-slope form is,

  $\begin{array}{l}y-1=\frac{9-1}{5-\left(-5\right)}\left(x-\left(-5\right)\right)\\ \Rightarrow y-1=\frac{8}{10}\left(x+5\right)\\ \Rightarrow y-1=\frac{4}{5}\left(x+5\right)\end{array}$

Now the equation of the line in slope-intercept form is,

  $\begin{array}{l}y-1=\frac{4}{5}\left(x+5\right)\\ \Rightarrow y-1=\frac{4}{5}x+4\\ \Rightarrow y=\frac{4}{5}x+4+1\\ \Rightarrow y=\frac{4}{5}x+5\end{array}$

Now the equation of the line in standard form is,

  $\begin{array}{l}y=\frac{4}{5}x+5\\ \Rightarrow y=\frac{4x+25}{5}\\ \Rightarrow 5y=4x+25\\ \Rightarrow 4x-5y=-25\end{array}$