# 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.

3x-2y=12

=> 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.