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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. - Mathematics and Statistics

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Question

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.

Sum

Solution

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 m+ m= `- "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 m.

∴ 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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General Second Degree Equation in x and y
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Chapter 4: Pair of Straight Lines - Exercise 4.3 [Page 127]

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