Question

Write in standard form of the line that passes through the point (2,-3). its perpendicular. 6x-8y-5=0

Answer 1

well remember a perpendicular line has the same y intercept....but the negative reciprocal of the slope....

so lets put the equation in slope intercept for a second

6x - 8y - 5 = 0
6x - 8y = 5

-8y = 5 - 6x

y = -5/8 + 6/8x

written correctly y = 6/8x - 5/8

Okay so thats our line we have.....now we have a point of the other line

use the point slope formula for this...knowing that the slope must be the negative resiprocal of this other line

so

y - (-3) = -8/6(x - 2)

y + 3 = -8/6x + 16/6

y = -8/6x + 16/6 - 3

y = -8/6x - .33333333

now check...using the point given....plug into this equation

-3 = -8/6(2) -.3333333

-3 = -3

check...so thats your equation for the second line

but thats in slope intercept form...and it asks for standard form so

y = -8/6x - .33333333

y + 8/6x + .3333333 = 0

p.s .333 could also be written as 1/3

so you have

y + 8/6x + 1/3 = 0

Hope I helped :)

Answer 2

thanks:p

Answer 3

Anytime :)