Question

The original line has the equation of y=-2x+8
Write a new equation that is perpendicular to the original that goes through the point (6,-1)

Answer 1

The new equation of line that is perpendicular to the original that goes through the point (6, -1) in slope intercept form is $y = \frac{1}{2}x - 4$

Solution:

Given that original line has the equation of y = -2x + 8

We have to write a new equation that is perpendicular to the original that goes through the point (6, -1)

Let us first find slope of original line

The slope intercept form of line is given as:

y = mx + c ------ eqn 1

Where "m" is the slope of line and "c" is the y - intercept

On comparing the slope intercept form and given original equation, we get "m = -2"

Thus slope of original line "m" = -2

We know that product of slope of a line and slope of line perpendicular to it are always -1

slope of original line x slope of line perpendicular to it = -1

$\begin{array}{l}{-2 \times \text { slope of line perpendicular to it }=-1} \\\\ {\text { slope of line perpendicular to it }=\frac{1}{2}}\end{array}$

Let us find equation of line with slope m = 1/2 and passes through point (6, - 1)

Substitute $m = \frac{1}{2}$ and (x, y) = (6, -1) in eqn 1

$-1 = \frac{1}{2} \times 6 + c\\\\-1 = 3 + c\\\\c = -4$

Thus the required equation of line is:

Substitute "c" = -4 and $m = \frac{1}{2}$ in eqn 1

$y = \frac{1}{2}x - 4$

Thus the equation of line perpendicular to original line is found