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Abby I.

asked • 10/24/20

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=?

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?

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