Write the equation of the line passing through the point (-4,5) and perpendicular to 3x-2y=12
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The equation of the line passing through the point (-4,5) and perpendicular to 3x-2y=12 has to be determined.
=> y = (3/2)x - 6
This is in the y = mx + c form where m is the slope. The product of the slope of two perpendicular lines is -1.
The slope of the required line is -2/3
As it passes thorugh (-4, 5)
(y - 5)/(x + 4) = -2/3
=> 3y - 15 = -2x - 8
=> 2x + 3y - 7 = 0
The equation of the required line is 2x + 3y - 7 = 0
(-4,5) and perpendicular to 3x-2y=12
First find the slope of your line:
2y = 3x - 12
y = (3/2)x - 6
so, slope is 3/2
Find the negative reciprocal of that slope: -2/3
Plug it into the point slope formula to get:
y - 5 = (-2/3)(x + 4)
This is an acceptable answer or you can simplify further into slope-intercept form.
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