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Question
Find the joint equation of the line passing through the point (3, 2), one of which is parallel to the line x - 2y = 2, and other is perpendicular to the line y = 3.
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Solution
Let L1 be the line passes through (3, 2) and parallel to the line x - 2y = 2 whose slope is `(-1)/-2 = 1/2`
∴ slope of the line L1 is `1/2`
∴ equation of the line L2 is
y - 2 =`1/2`(x - 3)
∴ 2y - 4 = x - 3
∴ x - 2y + 1 = 0
Let L2 be the line passes through (3, 2) and perpendicular to the line y = 3.
∴ equation of the line L2 is of the form x = a. Since L2 passes through (3, 2), 3 = a.
∴ equation of the line L2 is x = 3, i.e. x - 3 = 0
Hence, the equations of the required lines are
x - 2y + 1 = 0 and x - 3 = 0
∴ their joint equation is
(x - 2y + 1)(x - 3) = 0
∴ x2 - 2xy + x - 3x + 6y - 3 = 0
∴ x2 - 2xy - 2x + 6y - 3 = 0
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