Question

what is the equation of a line that is perpendicular to the line shown and goes through the point (3,-1)

Answer 1

For lines to be perpendicular, the slope of one of them must be the inverse negative of the other one. So for example given the lines:

$y=mx+b$$y=nx+c$

For them to be perpendicular their slopes must be:

$n=-\frac{1}{m}$

To determine the equation of a line perpendicular to the given one, the first step is to calculate the slope of the graphed line.

Using two points of the line, for example: (0,3) and (4,6)

$\begin{gathered} m=\frac{y_1-y_2}{x_1-x_2} \\ m=\frac{6-3}{4-0}=\frac{3}{4} \end{gathered}$

The slope of the graphed line is

$m=\frac{3}{4}$

Now the slope of the perpendicular line must be the inverse negative so that:

$\begin{gathered} n=-\frac{1}{m} \\ n=-\frac{4}{3} \end{gathered}$

Now that we know the slope of the perpendicular line, and is given that it crosses the point (3,-1) we can use the point slope form to determine its equation

$y-y_1=m(x-x_1)$

For our line

$\begin{gathered} y-(-1)=-\frac{4}{3}(x-3) \\ y+1=-\frac{4}{3}x+4 \\ y=-\frac{4}{3}x+4-1 \\ y=-\frac{4}{3}x+3 \end{gathered}$

The equation of a perpendicular line of the one shown in the graph rthat goes through point (3,-1) is

$y=-\frac{4}{3}x+3$