The equation, in general form, of the line that passes through the point (2,−11) and is perpendicular to the line 8x+7y+7=0 is Ax+By+C=0, where A= ?, B= ?, and C=?

Question

The equation, in general form, of the line that passes through the point (2,−11) and is perpendicular to the line 8x+7y+7=0 is Ax+By+C=0, where

A=

B=

C=

Here's how I've attempted the question so far:

Y=mx+ b

y-y1=m(x-x1)

y-(-11)=m(x-2)

x1= 2, y1=-11

8x + 7y +7= 0 rewritten is y= -8/7 x - 1

m = 7/8

y-(-11) = 7/8 (x-2)

11= 7/8x -7/4

-7/8

=81/8

I know that this is incorrect, please help me solve this and where my mistake was made?

Answer 1

8x+7y +7 =0, rewrite solving for y

7y= -8x-7

y = (-8/7)x -1

the slope is -8/7

the slope of the perpendicular line is the negative inverse, change the sign and flip it upside down: 7/8

the perpendicular equation is y=(7/8)x + constant

plug the point in to get the constant term

-11 = (7/8)(2) + c

-11 = 7/4 + c

c = -11 - 7/4 = -51/4

the perpendicular line is

y=(7/8)x -51/4 or

8y = 7x -102

I think your mistake was when you didn't plug in the point into the perpendicular equation

you should have gotten -11 = (7/8)(2) + constant,

you seem to have something different. you should just replace y with -11 and replace x with 2