Question 159144
    write an equation of the line containing the given point and perpendicular to the given line (8,-4); 6x+7y=2
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First, let's rewrite the given formula into the "slope-intercept" form of a line:
y = mx + b
where
m is slope
b is the y-intercept
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So, from:
6x+7y=2
7y=-6x+2
y = (-6/7)x + 2/7
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Now, we know the slope is -6/7
If a line is to be perpendicular, the slope has to be the negative reciprocal:
M(-6/7) = -1
M = 7/6
This is the slope of a perpendicular line.
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Plug our newly found slope, 7/6, and the given point (8,-4) into the "point-slope form" and solve:
y – y1 = m(x – x1)
y – (-4) = (7/6)(x – 8)
y + 4 = (7/6)x – (7/6)8
y + 4 = (7/6)x – (7/3)4
y + 4 = (7/6)x – (28/3)
y = (7/6)x – (28/3) - 4
y = (7/6)x – (28/3) - 12/3
y = (7/6)x – 40/3