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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). - Mathematics

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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).

योग
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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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अध्याय 10: Straight Lines - Exercise [पृष्ठ १७८]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
अध्याय 10 Straight Lines
Exercise | Q 2 | पृष्ठ १७८

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