If the equations 3x^2 + 2hxy - 3y^2 = 0 and 3 | vedclass

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Question

(A) 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

$\frac{1}{4}$

Vedclass · Answer

(A) 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 these to be the sides of a square,the lines must be perpendicular. Since the second equation $3x^2 + 2hxy - 3y^2 + 2x - 4y + c = 0$ represents the other two sides,they must be parallel to the first pair. Comparing the equations,the distance between the parallel lines $3x^2 + 2hxy - 3y^2 + 2x - 4y + c = 0$ and $3x^2 + 2hxy - 3y^2 = 0$ must be equal for both pairs. For the lines to form a square,the distance between the parallel lines must be equal. By solving for the condition of a square,we find $h = 2$. Substituting the values into the equation,we obtain $c = -8$. Thus,$\frac{h}{c} = \frac{2}{-8} = -\frac{1}{4}$. Given the options,the correct evaluation leads to $\frac{h}{c} = -\frac{1}{4}$.

If the equations $3x^2 + 2hxy - 3y^2 = 0$ and $3x^2 + 2hxy - 3y^2 + 2x - 4y + c = 0$ represent the four sides of a square,then $\frac{h}{c} =$

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