The joint equation of the pair of lines passi | vedclass

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(B) The joint equation of the pair of lines passing through the point $(x_1, y_1)$ and perpendicular to the lines $ax^2 + 2hxy + by^2 = 0$ is given by

Vedclass · Accepted Answer

$3x^2 + 2xy - 5y^2 - 14x - 26y - 5 = 0$

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(B) The joint equation of the pair of lines passing through the point $(x_1, y_1)$ and perpendicular to the lines $ax^2 + 2hxy + by^2 = 0$ is given by $b(x - x_1)^2 - 2h(x - x_1)(y - y_1) + a(y - y_1)^2 = 0$.Here,$a = 5$,$2h = 2$ (so $h = 1$),$b = -3$,and $(x_1, y_1) = (3, -2)$.Substituting these values into the formula:$-3(x - 3)^2 - 2(x - 3)(y + 2) + 5(y + 2)^2 = 0$Expanding the terms:$-3(x^2 - 6x + 9) - 2(xy + 2x - 3y - 6) + 5(y^2 + 4y + 4) = 0$$-3x^2 + 18x - 27 - 2xy - 4x + 6y + 12 + 5y^2 + 20y + 20 = 0$Combining like terms:$-3x^2 - 2xy + 5y^2 + 14x + 26y + 5 = 0$Multiplying by $-1$ to standardize:$3x^2 + 2xy - 5y^2 - 14x - 26y - 5 = 0$.

The joint equation of the pair of lines passing through the point $(3, -2)$ and perpendicular to the lines $5x^2 + 2xy - 3y^2 = 0$ is:

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