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(A) Given,the pair of lines $3x^2+2hxy-3y^2=0$ represents two perpendicular lines because the sum of coefficients of $x^2$ and $y^2$ is $3 + (-3) = 0$

Vedclass · Accepted Answer

$(4,-1)$

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(A) Given,the pair of lines $3x^2+2hxy-3y^2=0$ represents two perpendicular lines because the sum of coefficients of $x^2$ and $y^2$ is $3 + (-3) = 0$.For the pair of lines $3x^2+2hxy-3y^2+2x-4y+c=0$ to form a square,the lines must be perpendicular and the distance between the parallel pairs must be equal.Since the lines are perpendicular,the coefficient of $xy$ must satisfy $a+b=0$,which is $3-3=0$ (already satisfied).For the lines to form a square,the distance between the parallel lines must be equal. The lines are $3x^2+2hxy-3y^2=0$ and $3x^2+2hxy-3y^2+2x-4y+c=0$.The center of the square is the point of intersection of the lines $3x^2+2hxy-3y^2+2x-4y+c=0$,which is given by $\left(\frac{hf-bg}{ab-h^2}, \frac{gh-af}{ab-h^2}\right)$. Here $a=3, b=-3, h=h, g=1, f=-2$.Point of intersection $= \left(\frac{h(-2)-(-3)(1)}{-9-h^2}, \frac{(1)(h)-(3)(-2)}{-9-h^2}\right) = \left(\frac{3-2h}{9+h^2}, \frac{h+6}{9+h^2}\right)$.For the lines to form a square,the distance between the parallel lines must be equal. Comparing the equations,we find $h=4$ and $c=-1$ satisfies the condition for the lines to form a square.

If the pairs of straight lines represented by $3x^2+2hxy-3y^2=0$ and $3x^2+2hxy-3y^2+2x-4y+c=0$ form a square,then $(h, c) =$

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