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(B) The given equation 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 o

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$3x+2y-10=0$

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(B) The given equation 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 $(x_1, y_1)$ is given by $\left(\frac{bg-fh}{h^2-ab}, \frac{af-gh}{h^2-ab}\right)$. Substituting the values: $x_1 = \frac{3(-2)-(-5)(2)}{2^2-1(3)} = \frac{-6+10}{1} = 4$ and $y_1 = \frac{1(-5)-(-2)(2)}{2^2-1(3)} = \frac{-5+4}{1} = -1$. So,the point of intersection is $(4, -1)$. The equation of the line passing through $(4, -1)$ and $(2, 2)$ is given by $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$. $y - (-1) = \frac{2 - (-1)}{2 - 4}(x - 4) \Rightarrow y + 1 = \frac{3}{-2}(x - 4)$. $-2y - 2 = 3x - 12 \Rightarrow 3x + 2y - 10 = 0$.

The equation of the straight line passing through the point of intersection of the lines represented by $x^2+4xy+3y^2-4x-10y+3=0$ and the point $(2,2)$ is

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