Question

hi how do i , Find the equation of the line in slope intercept form that is perpendicular to the line 2x - 8y + 20 = 0 that goes through the point (-1, 6)

Answer 1

We have the following equation:

$2x-8y+20=0$

In order to locate the slope of this line equation, lets isolate the variable y. Then, by subtracting 2x and 20 to both sides, we have

$-8y=-2x-20$

and by dividing both sides by -8, we obtain

$\begin{gathered} y=\frac{-2x-20}{-8} \\ y=-\frac{2}{-8}x-\frac{20}{-8} \\ y=\frac{1}{4}x+\frac{5}{2} \end{gathered}$

By comparing this result with the slope-intercept form

$y=mx+b$

we can see that the slope m of the given line equation is

$m=\frac{1}{4}$

Since we need a perperdicular line, we need to compute the negative reciprocal of the slope m, that is,

$\begin{gathered} M\questeq-\frac{1}{m} \\ so \\ M=-\frac{1}{\frac{1}{4}}=-4 \end{gathered}$

Then, the perpendicular line has the form

$\begin{gathered} y=Mx+b \\ \text{that is, } \\ y=-4x+b \end{gathered}$

Finally, we can find the y-intercept b by substituting the coordinates of the given point (-1,6). It yields,

$6=-4(-1)+b$

which gives

$\begin{gathered} 6=4+b \\ \text{then} \\ b=2 \end{gathered}$

Therefore, the perpendicular line to 2x-8y+20=0 that goes throught the point (-1,6) is:

$y=-4x+2$