The point of intersection of the lines repres | vedclass

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Question

(C) The given equation is $x^2 - y^2 + x + 3y - 2 = 0$.Comparing this with the general second-degree equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$

Vedclass · Accepted Answer

$(-\frac{1}{2}, \frac{3}{2})$

Vedclass · Answer

(C) The given equation is $x^2 - y^2 + x + 3y - 2 = 0$.Comparing this with the general second-degree equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we get $a=1, h=0, b=-1, g=\frac{1}{2}, f=\frac{3}{2}, c=-2$.The point of intersection $(x, y)$ of the pair of lines is given by the solution of the partial derivatives $\frac{\partial f}{\partial x} = 0$ and $\frac{\partial f}{\partial y} = 0$.$\frac{\partial}{\partial x}(x^2 - y^2 + x + 3y - 2) = 2x + 1 = 0 \Rightarrow x = -\frac{1}{2}$.$\frac{\partial}{\partial y}(x^2 - y^2 + x + 3y - 2) = -2y + 3 = 0 \Rightarrow y = \frac{3}{2}$.Thus,the point of intersection is $(-\frac{1}{2}, \frac{3}{2})$.

The point of intersection of the lines represented by the equation $x^2 - y^2 + x + 3y - 2 = 0$ is

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