If (-1, -1) is the point of intersection of t | vedclass

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(D) The given equation represents a pair of lines: $f(x, y) = 2x^2 + 5xy - 3y^2 + 2gx + 2fy + c = 0$.
For a general second-degree equation $ax^2 + 2h

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) The given equation represents a pair of lines: $f(x, y) = 2x^2 + 5xy - 3y^2 + 2gx + 2fy + c = 0$.
For a general second-degree equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,the point of intersection $(x_0, y_0)$ is found by solving the partial derivatives $\frac{\partial f}{\partial x} = 0$ and $\frac{\partial f}{\partial y} = 0$.
$\frac{\partial f}{\partial x} = 4x + 5y + 2g = 0$.
Substituting the point of intersection $(-1, -1)$:
$4(-1) + 5(-1) + 2g = 0 \implies -4 - 5 + 2g = 0 \implies 2g = 9 \implies g = 4.5$.
$\frac{\partial f}{\partial y} = 5x - 6y + 2f = 0$.
Substituting the point of intersection $(-1, -1)$:
$5(-1) - 6(-1) + 2f = 0 \implies -5 + 6 + 2f = 0 \implies 1 + 2f = 0 \implies f = -0.5$.
Therefore,$g + f = 4.5 + (-0.5) = 4$.
Thus,the correct option is $D$.

If $(-1, -1)$ is the point of intersection of the pair of lines $2x^2 + 5xy - 3y^2 + 2gx + 2fy + c = 0$,then $g + f =$

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