Question

What is the equation of the line that is perpendicular to the line through (1,-4) and (-2,2) and passes through the x-intercept of that line?

Answer 1

Answer:

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

Step-by-step explanation:

First, calculate the slope of the line that is perpendicular to the equation of line we are asked to find

m=(y2-y1)/(x2-x1)

  =(2-(-4))/(-2-1)

  =6/-3

  =-2

in this equation the slope is 2, and to find the first equation, use y=mx+b

use the point (1, -4) to find b

-4=(2)(1)+b

-4=2+b

b=-6

the first equation of the line is y=2x-6

to find the x intercept of that line substitute 0 for y

0=2x-6

2x=6

x=3

the slope of a line perpendicular to this would be the opposite reciprocal of the slope which would be equal to -1/2

for the second equation of the line to pass thorugh the x-intercept of the first line, it must pass through (3, 0), so substitute and solve for b

y=mx+b

0=(-1/2)(3)+b

b=3/2

thus the equation of the line that is perpendicular to the line through (1,-4) and (-2, 2) and passes through the x intercept of that line is y=-$\frac{1}{2}$x+3/2