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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