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

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(C) The given equation 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

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$(-1, 1)$

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(C) The given equation 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 $c = -1$. The angle $	heta$ between the lines is given by $	an 	heta = |\frac{2\sqrt{h^2 - ab}}{a + b}|$. Given $	an 	heta = 3/4$,we have $3/4 = |\frac{2\sqrt{h^2 - 4}}{2 + 2}| = |\frac{2\sqrt{h^2 - 4}}{4}| = \frac{\sqrt{h^2 - 4}}{2}$. Squaring both sides,$9/16 = (h^2 - 4)/4$,which gives $9/4 = h^2 - 4$,so $h^2 = 25/4$,implying $h = 5/2$ (since $h > 0$). The point of intersection $(x, y)$ is found by solving $\frac{\partial f}{\partial x} = 0$ and $\frac{\partial f}{\partial y} = 0$. $\frac{\partial f}{\partial x} = 4x + 2hy - 1 = 0 \implies 4x + 5y - 1 = 0$. $\frac{\partial f}{\partial y} = 2hx + 4y + 1 = 0 \implies 5x + 4y + 1 = 0$. Solving these equations: $16x + 20y - 4 = 0$ and $25x + 20y + 5 = 0$. Subtracting gives $9x + 9 = 0$,so $x = -1$. Substituting $x = -1$ into $4x + 5y - 1 = 0$ gives $-4 + 5y - 1 = 0$,so $5y = 5$,$y = 1$. The point of intersection 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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