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Question
Find the equation of the line passing through (–3, 5) and perpendicular to the line through the points (2, 5) and (–3, 6).
Solution
Slope of the line passing through the points A(2, 5) and B(−3, 6)
`"m"_1 = ("y"_2 -"y"_1)/("x"_2 - "x"_1)`
= `(6 - 5)/(-3 -2)`
= `1/(-5)`
= `-1/5`
If PL is perpendicular to AB from the point P(–3, 5), then consider its slope as m2.
Lines PL and AB are mutually perpendicular.
If Slope of PL × Slope of AB = –1
In other words, m2 × `(-1/5)` = −1
∴ m2 = 5
The slope of PL is 5 and it passes through P(−3, 5), then the equation of PL is,
y – y1 = m2(x – x1)
or y – 5 = 5 (x + 3)
∴ 5x – y + 20 = 0
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