मराठी

Find the equation to the straight line passing through the point of intersection of the lines 5x – 6y – 1 = 0 and 3x + 2y + 5 = 0 and perpendicular to the line 3x – 5y + 11 = 0.

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प्रश्न

Find the equation to the straight line passing through the point of intersection of the lines 5x – 6y – 1 = 0 and 3x + 2y + 5 = 0 and perpendicular to the line 3x – 5y + 11 = 0.

बेरीज
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उत्तर

First we find the point of intersection of lines

5x – 6y – 1 = 0 and 3x + 2y + 5 = 0 which is (– 1, – 1).

Also the slope of the line 3x – 5y + 11 = 0 is `3/5`.

Therefore, the slope of the line perpendicular to this line is `(-5)/3`  (Why?).

Hence, the equation of the required line is given by

y + 1 = `(-5)/3 (x + 1)`

or 5x + 3y + 8 = 0

Alternatively: The equation of any line through the intersection of lines 5x – 6y – 1 = 0 and 3x + 2y + 5 = 0 is

5x – 6y – 1 + k(3x + 2y + 5) = 0  ....(1)

or Slope of this line is `(-(5 + 3k))/(-6 + 2k)`

Also, slope of the line 3x – 5y + 11 = 0 is `3/5`

Now, both are perpendicular

So `(-(5 + 3k))/(-6 + 2k) xx 3/5` = –1

or k = 45

Therefore, equation of required line in given by

5x – 6y – 1 + 45(3x + 2y + 5) = 0

or 5x + 3y + 8 = 0

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पाठ 10: Straight Lines - Solved Examples [पृष्ठ १७१]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 11
पाठ 10 Straight Lines
Solved Examples | Q 9 | पृष्ठ १७१

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