Question 638150: Write an equation of the line contianing the given point and perpendicular to the given line.
(0,4), 8x+7y=3
Answer by DrBeeee(684)   (Show Source):
You can put this solution on YOUR website! Rearrange your equation to the slope-intercept form
y = mx + b
Given
8x+7y=3
7y=-8x+3
(1) y = -(8/7)x + 3/7
The requirement for a line to be perpendicular to a line of slope m is
slope of new line is -1/m
In your line the slope m = -(8/7), so the new slpoe is
m' = 7/8
New eqution of a line perpendicular to (1) is
(2) y = (7/8)x + 3
To make this line go through (0,4), start with
(3) y = (7/8)x + b, where b is called the y-intercept
and put in the desired point (0,4). We get
4 = (7/8)(0) + b which simplifies to
b = 4
Your final equation is
(4) y = (7/8)x + 4
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