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प्रश्न
Find the joint equation of the pair of the line through the point (2, -3) and parallel to the lines represented by x2 + xy - y2 = 0.
उत्तर
The combined equation of the given lines is
x2 + xy - y2 = 0 ....(1)
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 (1).
Then m1 + m2 = `- "2h"/"b" = (-1)/-1 = 1` and m1m2 = `"a"/"b" = 1/-1 = - 1` .....(2)
The slopes of the lines parallel to these lines are m1 and m2 .
∴ the equations of the lines with these slopes and through the point (2, - 3) are
y + 3 = m1(x - 2) and y + 3 = m2 (x - 2)
i.e. m1(x - 2) - (y + 3) = 0 and m2(x - 2) - (y + 3) = 0
∴ the joint equation of these lines is
[m1 (x - 2) - (y + 3)][m2(x - 2)- (y + 3)] = 0
∴ m1m2 (x - 2)2 - m1 (x - 2)(y + 3) - m2(x - 2)(y+3) + (y + 3)2 = 0
∴ m1m2 (x - 2)2 - (m1 + m2)(x - 2)(y + 3) + (y + 3)3 = 0
∴ `- ("x" - 2)^2 - ("x" - 2)("y" + 3) + ("y" + 3)^2 = 0` ....[By (2)]
∴ `("x" - 2)^2 + ("x" - 2)("y" + 3) - ("y" + 3)^2 = 0`
∴ `("x"^2 - "4x" + 4) + ("xy" + "3x" - "2y" - 6) - ("y"^2 + "6y" + 9) = 0`
∴ `"x"^2 - "4x" + 4 + "xy" + "3x" - "2y" - 6 - "y"^2 - "6y" - 9 = 0`
∴ `"x"^2 + "xy" - "y"^2 - "x" - "8y" - 11 = 0`
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