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प्रश्न
Find the equation of the line passing through the point (5, 2) and perpendicular to the line joining the points (2, 3) and (3, – 1).
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उत्तर
Slope of the line joining the points (2, 3) and (3, – 1) is
`(-1 - 3)/(3 - 2)` = – 4
Slope of the required line which is perpendicular to it
= `(-1)/(-4)`
= `1/4` ....[m1m2 = – 1]
Equation of the line passing through the point (5, 2) is
y – 2 = `1/4(x - 5)` .....[y – y1 = m(x – x1)]
⇒ 4y – 8 = x – 5
⇒ x – 4y + 3 = 0
Hence, the required equation is x – 4y + 3 = 0.
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| Column C1 | Column C2 |
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