Let PQR be a right-angled isosceles triangle, | vedclass

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(A) Given that $PQR$ is a right-angled isosceles triangle at $Q(2, 1)$. The line $PR$ has the equation $2x + y = 3$,so its slope is $m = -2$.Since $

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$3x^2 + 8xy - 3y^2 - 20x - 10y + 25 = 0$

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(A) Given that $PQR$ is a right-angled isosceles triangle at $Q(2, 1)$. The line $PR$ has the equation $2x + y = 3$,so its slope is $m = -2$.Since $	riangle PQR$ is isosceles and right-angled at $Q$,the lines $PQ$ and $QR$ make an angle of $45^\circ$ with the line $PR$.Let the slope of $PQ$ be $m_1$. Then,$	an 45^\circ = |\frac{m_1 - (-2)}{1 + m_1(-2)}| = |\frac{m_1 + 2}{1 - 2m_1}|$.$1 = |\frac{m_1 + 2}{1 - 2m_1}| \Rightarrow 1 - 2m_1 = m_1 + 2$ or $1 - 2m_1 = -(m_1 + 2)$.Case $1$: $3m_1 = -1 \Rightarrow m_1 = -1/3$.Case $2$: $m_1 = 3$.Thus,the slopes of $PQ$ and $QR$ are $-1/3$ and $3$.The equations of the lines passing through $Q(2, 1)$ are:$y - 1 = -1/3(x - 2)$ $\Rightarrow 3y - 3 = -x + 2$ $\Rightarrow x + 3y - 5 = 0$.$y - 1 = 3(x - 2)$ $\Rightarrow y - 1 = 3x - 6$ $\Rightarrow 3x - y - 5 = 0$.The combined equation is $(x + 3y - 5)(3x - y - 5) = 0$.Expanding this: $3x^2 - xy - 5x + 9xy - 3y^2 - 15y - 15x + 5y + 25 = 0$.$3x^2 + 8xy - 3y^2 - 20x - 10y + 25 = 0$.

Let $PQR$ be a right-angled isosceles triangle,with the right angle at $Q(2, 1)$. If the equation of the line $PR$ is $2x + y = 3$,then the combined equation representing the pair of lines $PQ$ and $QR$ is:

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