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Question
Find the joint equation of the pair of a line through the origin and perpendicular to the lines given by
x2 + xy - y2 = 0
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Solution
Comparing the equation x2 + xy - y2 = 0 with ax2 + 2hxy + by2 = 0, we get,
a = 1, 2h = 1, b = - 1
Let m1 and m2 be the slopes of the lines represented by x2 + xy - y2 = 0
∴ m1 + m2 = `(-"2h")/"b" = (-1)/-1 = 1` and m1m2 = `"a"/"b" = 1/-1 = -1` ...(1)
Now, required lines are perpendicular to these lines
∴ their slopes are `(-1)/"m"_1` and `- 1/"m"_2`
Since these lines are passing through the origin, their separate equations are
y = `(-1)/"m"_1 "x"` and y = `(-1)/"m"_2 "x"`
i.e. m1y = - x and m2y = - x
i.e. x + m1y = 0 and x + m2y = 0
∴ their combined equation is
(x + m1y)(x + m2y) = 0
∴ x2 + (m1 + m2)xy + m1m2y2 = 0
∴ `"x"^2 + 1"xy" + (-1)"y"^2 = 0` ....[By(1)]
∴ `"x"^2 + "xy" - "y"^2 = 0`
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