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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).
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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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