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Question
Find the joint equation of the line passing through (-1, 2) and perpendicular to the lines x + 2y + 3 = 0 and 3x - 4y - 5 = 0
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Solution
Let L1 and L2 be the lines passing through the origin and perpendicular to the lines x + 2y + 3 = 0 and 3x - 4y - 5 = 0 respectively.
Slopes of the lines x + 2y + 3 = 0 and 3x - 4y - 5 = 0 are `-1/2` and `- 3/-4 = 3/4` respectively.
∴ slopes of the lines L1 and L2 are `2` and `(-4)/3` respectively.
Since the lines L1 and L2 pass through the point (-1, 2), their equations are
∴ (y - y1) = m(x - x1)
∴ (y - 2)= 2(x + 1)
⇒ y - 2 = 2x + 2
⇒ 2x - y + 4 = 0 and
∴ (y - 2) = `((-4)/3)`(x + 1)
⇒ 3y - 6 = (- 4)(x + 1)
⇒ 3y - 6 = - 4x - 4
⇒ 4x + 3y - 6 + 4 = 0
⇒ 4x + 3y - 2 = 0
their combined equation is
∴ (2x - y + 4)(4x + 3y - 2) = 0
∴ 8x2 + 6xy - 4x - 4xy - 3y2 + 2y + 16x + 12y - 8 = 0
∴ 8x2 + 2xy + 12x - 3y2 + 14y - 8 = 0
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