Question

What is an equation of the line that is perpendicular to y-4=2(x-6) and passes through the point (-3,-5)

Answer 1

Answer:

$y+5=-\frac{1}{2}(x+3)$

Step-by-step explanation:

The given equation is

$y-4=2(x-6)$

This equation is in the point-slope form, $y-y_1=m(x-x_1)$, where $m=2$ is the slope of this line.

The line that is perpendicular to this line has slope $-\frac{1}{2}$

If this line that is perpendicular to the given line passes through (-3,-5), then it will have equation;

$y--5=-\frac{1}{2}(x--3)$

$y+5=-\frac{1}{2}(x+3)$ in point-slope form

Answer 2

Answer:

y+5 = -1/2(x + 3)

Step-by-step explanation:

Perpendicular lines have negative reciprocal slopes. The slope of the line

y - 4 = 2(x - 6) is 2. Its negative reciprocal slope is -1/2.

Using m = -1/2 and the point (-3,-5) in the point slope form.

y --5 = -1/2(x--3)

y+5 = -1/2(x + 3)