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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.
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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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