If the equation kx^2 - 5xy + 6y^2 + x - 3y = | vedclass

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(A) The general equation of a pair of lines is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$. Comparing this with $kx^2 - 5xy + 6y^2 + x - 3y = 0$, we get

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$(-3, -1)$

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(A) The general equation of a pair of lines is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$. Comparing this with $kx^2 - 5xy + 6y^2 + x - 3y = 0$, we get $a=k, h=-5/2, b=6, g=1/2, f=-3/2, c=0$.For the lines to represent a pair of straight lines, the condition is $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.Substituting the values: $k(6)(0) + 2(-3/2)(1/2)(-5/2) - k(-3/2)^2 - 6(1/2)^2 - 0(-5/2)^2 = 0$.$0 + 15/4 - 9k/4 - 6/4 = 0 \implies 9/4 - 9k/4 = 0 \implies k=1$.The equation becomes $x^2 - 5xy + 6y^2 + x - 3y = 0$.The point of intersection $(x, y)$ is found by partial differentiation with respect to $x$ and $y$: $\frac{\partial}{\partial x} = 2x - 5y + 1 = 0$ and $\frac{\partial}{\partial y} = -5x + 12y - 3 = 0$.Solving these: $2x - 5y = -1$ $(i)$ and $-5x + 12y = 3$ (ii).Multiply $(i)$ by $5$ and (ii) by $2$: $10x - 25y = -5$ and $-10x + 24y = 6$.Adding them: $-y = 1 \implies y = -1$.Substitute $y = -1$ in $(i)$: $2x - 5(-1) = -1 \implies 2x + 5 = -1 \implies 2x = -6 \implies x = -3$.The point of intersection is $(-3, -1)$.

If the equation $kx^2 - 5xy + 6y^2 + x - 3y = 0$ represents a pair of straight lines, then the coordinates of their point of intersection are

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