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Question
Find the equation of the pair of straight lines passing through the point (1, 3) and perpendicular to the lines 2x − 3y + 1 = 0 and 5x + y − 3 = 0
Solution
Equation of a line perpendicular to 2x – 3y + 1 = 0 is of the form 3x + 2y + k = 0.
It passes through (1, 3)
⇒ 3 + 6 + k = 0
⇒ k = – 9
So the line is 3x + 2y – 9 = 0
The equation of a line perpendicular to 5x + y – 3 = 0 will be of the form x – 5y + k = 0.
It passes through (1, 3)
⇒ 1 – 15 + k = 0
⇒ k = 14
So the line is x – 5y + 14 = 0.
The equation of the lines is 3x + 2y – 9 = 0 and x – 5y + 14 = 0
Their combined equation is (3x + 2y – 9)(x – 5y + 14) = 0
(i.e) 3x2 – 15xy + 42x + 2xy – 10y2 + 28y – 9x + 45y – 126 = 0
(i.e) 3x2 – 13xy – 10y2 + 33x + 73y – 126 = 0
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