If the angle between the pair of lines 2x^2 + | vedclass

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(C) The given equation of the pair of lines is $2x^2 + 2hxy + 2y^2 - x + y - 1 = 0$. Comparing this with the general form $ax^2 + 2h'xy + by^2 + 2gx +

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$(-1, 1)$

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(C) The given equation of the pair of lines is $2x^2 + 2hxy + 2y^2 - x + y - 1 = 0$. Comparing this with the general form $ax^2 + 2h'xy + by^2 + 2gx + 2fy + c = 0$,we have $a = 2$,$h' = h$,$b = 2$,$g = -1/2$,$f = 1/2$,and $c = -1$.For a pair of lines to intersect at a point $(x_0, y_0)$,the partial derivatives with respect to $x$ and $y$ must be zero:$\frac{\partial}{\partial x}(2x^2 + 2hxy + 2y^2 - x + y - 1) = 4x + 2hy - 1 = 0$$\frac{\partial}{\partial y}(2x^2 + 2hxy + 2y^2 - x + y - 1) = 2hx + 4y + 1 = 0$Solving these equations for $x$ and $y$:$4x + 2hy = 1$ and $2hx + 4y = -1$.Using Cramer's rule or substitution,the determinant of the system is $D = 16 - 4h^2$. For the lines to exist,$D 
eq 0$. The intersection point is $(x_0, y_0) = (\frac{2+h}{8-2h^2}, \frac{-2-h}{8-2h^2}) = (\frac{2+h}{2(4-h^2)}, \frac{-(2+h)}{2(4-h^2)}) = (\frac{1}{2(2-h)}, \frac{-1}{2(2-h)})$.The angle $	heta$ between the lines is given by $	an 	heta = |\frac{2\sqrt{h^2 - ab}}{a+b}| = |\frac{2\sqrt{h^2 - 4}}{2+2}| = |\frac{\sqrt{h^2 - 4}}{2}|$.Given $	an 	heta = 3/4$,we have $\frac{\sqrt{h^2 - 4}}{2} = 3/4 \implies \sqrt{h^2 - 4} = 3/2 \implies h^2 - 4 = 9/4 \implies h^2 = 25/4 \implies h = 5/2$.Substituting $h = 5/2$ into the intersection point: $x_0 = \frac{1}{2(2 - 5/2)} = \frac{1}{2(-1/2)} = -1$ and $y_0 = \frac{-1}{2(2 - 5/2)} = 1$. Thus,the point is $(-1, 1)$.

If the angle between the pair of lines $2x^2 + 2hxy + 2y^2 - x + y - 1 = 0$ is $\tan^{-1}(3/4)$ and $h$ is a positive rational number,then the point of intersection of these two lines is

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