Prove that the line joining the points $(4, 5)$ and $(0, - 2)$ is perpendicular to the line passing through $(2, - 3)$ and $(- 5, 1)$ .

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Solution

For line $1$ : Let $(x_{1}, y_{1}) = (4, 5)$ and $(x_{2}, y_{2}) = (0, - 2)$ .
Slope $= \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{- 2 - 5}{0 - 4}$
Slope $= \frac{- 7}{- 4} = \frac{7}{4} = m_{1}$
For line $2 : (x_{1}, y_{2}) = (2, - 3), (x_{2}, y_{2}) = (- 5, 1)$
Slope $= \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{1 - (- 3)}{- 5 - 2} = \frac{1 + 3}{- 7}$
Slope $= \frac{4}{7} = \frac{- 4}{7} = m_{2}$
$m_{1} \times m_{2} = \frac{7}{4} \times \frac{- 4}{7} = - 1$
The product of their slopes is $- 1$
$∴$ The two lines are perpendicular.

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Prove that the line joining the points (4,5) and (0,−2) is perpendicular to the line passing through (2,−3) and (−5,1).

Prove that the line joining the points $(4, 5)$ and $(0, - 2)$ is perpendicular to the line passing through $(2, - 3)$ and $(- 5, 1)$ .