How do you write an equation of a line passing through (-3, -2), perpendicular to 2x - 3y = 32x−3y=3?

1 Answer
EZ as pi
Aug 10, 2016

y = -3/2x -6 1/2y=−32x−612

Explanation:

We need the slope of the line we have been given before we can find the slope of the new line.

Change 2x-3y=32x−3y=3 into the form y = mx + cy=mx+c

2x -3 = 3y" "rArr" " 3y =2x-32x−3=3y ⇒ 3y=2x−3
y = 2/3x -1. " "m_1 = 2/3y=23x−1. m1=23.

The slope perpendicular to this is m_2 = -3/2m2=−32
We have the slope , (-3/2)(−32)and a point, (-3,-2)(−3,−2)
substitute the values into

y-y_1 = m(x-x_1)y−y1=m(x−x1)

y-(-2) = -3/2(x-(-3))y−(−2)=−32(x−(−3))

y+2 = -3/2(x+3)y+2=−32(x+3)

y= -3/2x-9/2-2y=−32x−92−2

y = -3/2x -6 1/2y=−32x−612

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