Question:
line through the points (-2, 6) and (4, 8) is perpendicular to the line through the points (3, -3) and (5, -9).
Solution:
For two lines to be perpendicular, their product of slope must be equal to -1.
Given points are A(-2,6),B(4,8) and C(3,-3),D(5,-9)
slope $=\left(\frac{\mathrm{y}_{2}-\mathrm{y}_{1}}{\mathrm{x}_{2}-\mathrm{x}_{1}}\right)$
Slope of line AB×slope of line CD = -1
$\Rightarrow\left(\frac{8-6}{4+2}\right) \times\left(\frac{-9+3}{5-3}\right)=-1$
$\Rightarrow\left(\frac{2}{6}\right) \times\left(\frac{-6}{2}\right)=-1 \Rightarrow-1=-1$
$\Rightarrow \mathrm{LHS}=\mathrm{RHS}$
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