Find the slope-intercept form for the line perpendicular to 2x –3y=-6 and passing through (4,–9)
For perpendicular lines, m1m2=−1m_1m_2 = -1m1m2=−1
From 2x−3y=−62x-3y=-62x−3y=−6 , we have that y=23x+2y=\frac{2}{3}x+2y=32x+2
Let m1=23m_1=\frac{2}{3}m1=32
=> m2=−32m_2=\frac{-3}{2}m2=2−3
m2=y−y1x−x1m_2=\frac{y-y_1}{x-x_1}m2=x−x1y−y1
−32=y−(−9)x−4=y+9x−4\frac{-3}{2}=\frac{y-(-9)}{x-4}=\frac{y+9}{x-4}2−3=x−4y−(−9)=x−4y+9
=> −3(x−4)=2(y+9)-3(x-4)=2(y+9)−3(x−4)=2(y+9)
12−3x=2y+1812-3x=2y+1812−3x=2y+18
=>y=−32x−3=> y = \frac{-3}{2}x-3=>y=2−3x−3
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