Chapter 4.2, Problem 26PPS

To determine

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

Answer to Problem 26PPS

Equation of the line in point-slope form is $y-11=3\left(x+4\right)$ .

Equation of the line in slope-intercept form is $y=3x+23$ .

Equation of the line in standard form is $3x-y=-23$

Explanation of Solution

Given:

Line passing through the points $\left(-4,11\right),\left(-8,-1\right)$ .

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,

  $\left(x_1,y_1\right)=\left(-4,11\right)\&\left(x_2,y_2\right)=\left(-8,-1\right)$

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

  $\begin{array}{l}y-11=\frac{-1-11}{-8-\left(-4\right)}\left(x-\left(-4\right)\right)\\ \Rightarrow y-11=\frac{-12}{-4}\left(x+4\right)\\ \Rightarrow y-11=3\left(x+4\right)\end{array}$

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

  $\begin{array}{l}y-11=3\left(x+4\right)\\ \Rightarrow y-11=3x+12\\ \Rightarrow y=3x+12+11\\ \Rightarrow y=3x+23\end{array}$

Now the equation of the line in standard form is,

  $\begin{array}{l}y=3x+23\\ \Rightarrow 3x-y=-23\end{array}$