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Question 178446: Write the equation of a line perpendicular to 7x +8y-12=0,which passes through the point (-2,5)
Answer by nerdybill(7384)   (Show Source):
You can put this solution on YOUR website! First, determine the slope of:
7x +8y-12=0
Do this by putting it into the "slope-intercept" form:
y = mx + b
where
m is slope
b is y-intercept
.
7x +8y-12=0
7x +8y= 12
8y= -7x + 12
y= (-7/8)x + 12/8
y= (-7/8)x + 3/2
So, now we know that the slope is -7/8
.
Our new slope has to be perpendicular (negative reciprocal):
(-7/8)m = -1
m = 8/7 (this is our new slope)
Our new slope along with the point provided (-2,5) can be plugged into the "point-slope" form:
y - y1 = m(x-x1)
y - 5 = (8/7)(x-(-2))
y - 5 = (8/7)(x+2)
y - 5 = (8/7)x + (8/7)2
y - 5 = (8/7)x + (16/7)
y = (8/7)x + (16/7) +5
y = (8/7)x + (16/7) + 35/7
y = (8/7)x + (51/7) (this is what they're looking for)
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