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

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(B) Let the slopes of $PQ$ and $PR$ be $m_1$ and $m_2$. Since $	riangle PQR$ is a right-angled isosceles triangle at $P$,the lines $PQ$ and $PR$ are

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

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(B) Let the slopes of $PQ$ and $PR$ be $m_1$ and $m_2$. Since $	riangle PQR$ is a right-angled isosceles triangle at $P$,the lines $PQ$ and $PR$ are perpendicular,so $m_1 m_2 = -1$. Also,the angle between $PQ$ and $QR$ is $45^\circ$. The slope of $QR$ is $m = -2$. Using the formula $	an 45^\circ = |\frac{m_1 - (-2)}{1 + m_1(-2)}| = 1$,we get $|\frac{m_1 + 2}{1 - 2m_1}| = 1$. This gives $m_1 + 2 = 1 - 2m_1$ or $m_1 + 2 = -(1 - 2m_1)$. Solving $3m_1 = -1$ gives $m_1 = -1/3$,and $m_1 + 2 = -1 + 2m_1$ gives $m_1 = 3$. Thus,the slopes are $3$ and $-1/3$. The equations of the lines passing through $P(2, 1)$ are $y - 1 = 3(x - 2)$ and $y - 1 = -1/3(x - 2)$. Rewriting these as $3x - y - 5 = 0$ and $x + 3y - 5 = 0$. The combined equation is $(3x - y - 5)(x + 3y - 5) = 0$. Expanding this: $3x^2 + 9xy - 15x - xy - 3y^2 + 5y - 5x - 15y + 25 = 0$,which simplifies to $3x^2 - 3y^2 + 8xy - 20x - 10y + 25 = 0$.

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

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