Find an equation for the line with the given properties. Perpendicular to the line x-7y=7 and contains the point (2,2).

Question

Please write your answer in slope-intercept form. Thank you.

Answer 1

x-7y = 7

solve for y and the slope = the coefficient of the x term: 1/7

perpendicular line has slope -7. take the -1/7 flip it upside down and change its sign: --7/1 = +7

y=mx+ b

y=7x +b, plug in the point (2,2) to solve for b. x=2, y=2

2 =7(2) +b

b= 2-14 =-12

perpendicular line is

y =7x -12

or in standard form

7x-y = 12

since the problem gave you the equation in standard form, the answer might be expected to be that way too

it helps to graph the lines and point

further common follow parts of the problem may be to find where the 2 lines intersect and what is the distance from that intersection to the given point= shortest distance from the point to the line

y = -x/7 -7= 7x-12

7x+x/7 = 12-7

50x/7 = 5

10x/7 = 1

10x = 7

x = 7/10=.7

y = -.7/7 = .1 = 1/10

the point (7/10, 1/10) is the point on the line closest to (2,2)

use the distance formula to find that distance

d^2 = (2-.7)^2 + (2-.1)^2 = 1.3^2 + 1.9^2 = 1.56 + 3.61

d = about 2.3022

Answer 2

x - 7y = 7

-7y - -x + 7

y = (1/7)x - 1, slope is 1/7

Slope of perpendicular is -7

2 = -7(x) + b

Can you sove for b and anser?