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(D) The given pair of lines is $x^2+4xy+3y^2-4x-10y+3=0$.Comparing with $ax^2+2hxy+by^2+2gx+2fy+c=0$,we have $a=1, h=2, b=3, g=-2, f=-5, c=3$.The poin

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$x^2+xy-6y^2-7x-16y+6=0$

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(D) The given pair of lines is $x^2+4xy+3y^2-4x-10y+3=0$.Comparing with $ax^2+2hxy+by^2+2gx+2fy+c=0$,we have $a=1, h=2, b=3, g=-2, f=-5, c=3$.The point of intersection $(x_0, y_0)$ is given by $\left(\frac{bg-fh}{h^2-ab}, \frac{af-gh}{h^2-ab}\right)$.Substituting the values: $x_0 = \frac{(3)(-2)-(-5)(2)}{2^2-(1)(3)} = \frac{-6+10}{4-3} = 4$ and $y_0 = \frac{(1)(-5)-(-2)(2)}{4-3} = \frac{-5+4}{1} = -1$.The point of intersection is $(4, -1)$.The equation of the line with slope $m_1 = \frac{1}{2}$ passing through $(4, -1)$ is $y+1 = \frac{1}{2}(x-4)$ $\Rightarrow 2y+2 = x-4$ $\Rightarrow x-2y-6=0$.The equation of the line with slope $m_2 = -\frac{1}{3}$ passing through $(4, -1)$ is $y+1 = -\frac{1}{3}(x-4)$ $\Rightarrow 3y+3 = -x+4$ $\Rightarrow x+3y-1=0$.The combined equation is $(x-2y-6)(x+3y-1) = 0$.Expanding this: $x(x+3y-1) - 2y(x+3y-1) - 6(x+3y-1) = 0$.$x^2+3xy-x-2xy-6y^2+2y-6x-18y+6 = 0$.$x^2+xy-6y^2-7x-16y+6=0$.

The combined equation of the pair of straight lines passing through the point of intersection of the pair of lines $x^2+4xy+3y^2-4x-10y+3=0$ and having slopes $\frac{1}{2}$ and $-\frac{1}{3}$ is

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