Question

I am trying to find the slope-intercept form of the following equation:Find the equation of a line through (7, -3) that is perpendicular to the line y = -x/3 - 8

Answer 1

Given:

The equation of line is,

$y=-\frac{x}{3}-8$

As given that the required line is perpendicular to the above line.

It means the slope of required line will be negative reciprocal of above line.

$\begin{gathered} y=-\frac{x}{3}-8 \\ \text{Compare it with y=mx+b} \\ \Rightarrow m=-\frac{1}{3} \end{gathered}$

The slope of the required line will be,

$m_1=-\frac{1}{m}=-\frac{1}{-\frac{1}{3}}=3$

Now, given that the required line passing through point (7,-3) .

$\begin{gathered} y=m_1x+b \\ (x,y)=(7,-3) \\ -3=3(7)+b \\ b=-24 \end{gathered}$

The slope-intercept form is,

$\begin{gathered} y=m_1x+b \\ y=3x-24 \end{gathered}$