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A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1:n. Find the equation of the line.

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प्रश्न

A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1:n. Find the equation of the line.

योग
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उत्तर

Line AB passes through the points A(1, 0) and B(2, 3).

∴ Slope of AB = `(3 - 0)/(2 - 1) = 3/1`

PQ ⊥ AB

Slope of AB = `3/1`

∴ Slope of PQ, m = `- 1/(3/1) = -1/3`

PQ intersects line AB at C.

Also, point C divides the line segment AB in the ratio 1 : n.

That is `"C"((1 xx 2 + "n" xx 1)/("n" + 1), (1 xx 3 + "n" xx 0)/("n" + 1))`

or `"C" (("n" + 2)/("n" + 1), 3/("n" + 1))`

Now the equation of line PQ,

`"y" - "y"_1 = "m"("x" - "x"_1)`

Where, x1 = `("n" + 2)/("n" + 1)` and y1 = `3/("n" + 1)`

`"y" - 3/("n" + 1) = -1/3("x" - ("n" + 2)/("n" + 1))`

3(n + 1)y − 9 = −[(n + 1)x − (n + 2)]

or (n + 1)x + 3(n + 1)y = n + 2 + 9 = n + 11

or (n + 1)x + 3(n + 1)y = n + 11

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अध्याय 9: Straight Lines - EXERCISE 9.2 [पृष्ठ १६४]

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एनसीईआरटी Mathematics [English] Class 11
अध्याय 9 Straight Lines
EXERCISE 9.2 | Q 10. | पृष्ठ १६४

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