The point of intersection of the lines repres | vedclass

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(C) The given equation is $2x^2 + 7xy + 3y^2 + 8x + 14y + 8 = 0$.Comparing this with the general equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we

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$(-2, 0)$

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(C) The given equation is $2x^2 + 7xy + 3y^2 + 8x + 14y + 8 = 0$.Comparing this with the general equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we get:$a = 2, b = 3, 2h = 7 \implies h = 3.5, 2g = 8 \implies g = 4, 2f = 14 \implies f = 7, c = 8$.The point of intersection $(x, y)$ is given by the formulas:$x = \frac{hf - bg}{ab - h^2} = \frac{(3.5)(7) - (3)(4)}{(2)(3) - (3.5)^2} = \frac{24.5 - 12}{6 - 12.25} = \frac{12.5}{-6.25} = -2$.$y = \frac{hg - af}{ab - h^2} = \frac{(3.5)(4) - (2)(7)}{(2)(3) - (3.5)^2} = \frac{14 - 14}{6 - 12.25} = 0$.Thus,the point of intersection is $(-2, 0)$.

The point of intersection of the lines represented by the equation $2x^2 + 7xy + 3y^2 + 8x + 14y + 8 = 0$ is

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