The joint equation of the pair of lines passi | vedclass

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Question

(A) The given equation is $x^{2} - 4xy + 3y^{2} = 0$. Factoring the equation: $x^{2} - 3xy - xy + 3y^{2} = 0$ $\Rightarrow x(x - 3y) - y(x - 3y) = 0$

Vedclass · Accepted Answer

$x^{2} + 3y^{2} - 4xy - 14x + 24y + 45 = 0$

Vedclass · Answer

(A) The given equation is $x^{2} - 4xy + 3y^{2} = 0$. Factoring the equation: $x^{2} - 3xy - xy + 3y^{2} = 0$ $\Rightarrow x(x - 3y) - y(x - 3y) = 0$ $\Rightarrow (x - y)(x - 3y) = 0$. The lines are $x - y = 0$ and $x - 3y = 0$. Lines parallel to these passing through $(3, -2)$ are of the form $(x - y + k_{1}) = 0$ and $(x - 3y + k_{2}) = 0$. For $(x - y + k_{1}) = 0$ passing through $(3, -2)$: $3 - (-2) + k_{1} = 0$ $\Rightarrow 5 + k_{1} = 0$ $\Rightarrow k_{1} = -5$. For $(x - 3y + k_{2}) = 0$ passing through $(3, -2)$: $3 - 3(-2) + k_{2} = 0$ $\Rightarrow 3 + 6 + k_{2} = 0$ $\Rightarrow k_{2} = -9$. The joint equation is $(x - y - 5)(x - 3y - 9) = 0$. Expanding: $x(x - 3y - 9) - y(x - 3y - 9) - 5(x - 3y - 9) = 0$. $x^{2} - 3xy - 9x - xy + 3y^{2} + 9y - 5x + 15y + 45 = 0$. $x^{2} + 3y^{2} - 4xy - 14x + 24y + 45 = 0$.

The joint equation of the pair of lines passing through $(3, -2)$ and parallel to the lines represented by $x^{2} - 4xy + 3y^{2} = 0$ is:

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