Show that the line joining $(2, - 5)$ and $(- 2, 5)$ is perpendicular to the line joining $(6, 3)$ and $(1, 1)$ .

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Solution

Slope of $(2, - 5)$ and $(- 2, 5) = \frac{5 + 5}{- 2 - 2} = \frac{10}{- 4} = \frac{5}{2} = m_{1}$

Slope of $(6, 3)$ and $(1, 1) = \frac{3 - 1}{6 - 1} = \frac{2}{5} = m_{2}$

$m_{1},_{2} = - (\frac{5}{2}) (\frac{2}{5}) = - 1$

Hence the given lines are perpendicular.

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Show that the line joining (2,−5) and (−2,5) is perpendicular to the line joining (6,3) and (1,1) .

Slope of (2,−5) and (−2,5)=5+5−2−2=10−4=52=m1Slope of (6,3) and (1,1)=3−16−1=25=m2m1,2=−(52)(25)=−1Hence the given lines are perpendicular.

Show that the line joining $(2, - 5)$ and $(- 2, 5)$ is perpendicular to the line joining $(6, 3)$ and $(1, 1)$ .

Slope of $(2, - 5)$ and $(- 2, 5) = \frac{5 + 5}{- 2 - 2} = \frac{10}{- 4} = \frac{5}{2} = m_{1}$

Slope of $(6, 3)$ and $(1, 1) = \frac{3 - 1}{6 - 1} = \frac{2}{5} = m_{2}$

$m_{1},_{2} = - (\frac{5}{2}) (\frac{2}{5}) = - 1$

Hence the given lines are perpendicular.