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Question

Find the values of $$p$$ and $$q$$, if the following equation represents a pair of perpendicular lines :
$$px^{2}-8xy+3y^{2}+14x+2y+q=0$$


Solution

We have,
$$px^2-8xy+3y^2+14x+2y+q=0$$

For perpendicular lines,
$$a+b=0$$
$$p+3=0$$
$$p=-3$$

Therefore,
$$-3x^2-8xy+3y^2+14x+2y+q=0$$

$$\left| \begin{matrix} a & h & g \\ h & b & f \\ g & f & c \end{matrix} \right| =0$$
$$\left| \begin{matrix} -3 & -4 & 7 \\ -4 & 3 & 1 \\ 7 & 1 & q \end{matrix} \right| =0$$
$$-3(3q-1)+4(-4q-7)+7(-4-21)=0$$
$$-9q+3-16q-28-28-147=0$$
$$25q=-200$$
$$q=-8$$

Hence, this is the answer.

Maths

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