Let PQR be an isosceles right-angled triangle | vedclass

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(B) Let the slopes of lines $PQ$ and $PR$ be $m_1$ and $m_2$. Since $\angle P = 90^\circ$,$m_1 m_2 = -1$. Also,since it is an isosceles triangle,the a

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

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(B) Let the slopes of lines $PQ$ and $PR$ be $m_1$ and $m_2$. Since $\angle P = 90^\circ$,$m_1 m_2 = -1$. Also,since it is an isosceles triangle,the angle between $PQ$ and $QR$ is $45^\circ$. The slope of $QR$ is $m = -2$.Using the formula $	an 	heta = |\frac{m_1 - m_2}{1 + m_1 m_2}|$,where $	heta = 45^\circ$:$1 = |\frac{m_1 - (-2)}{1 + m_1(-2)}| = |\frac{m_1 + 2}{1 - 2m_1}|$$(1 - 2m_1)^2 = (m_1 + 2)^2$$1 - 4m_1 + 4m_1^2 = m_1^2 + 4m_1 + 4$$3m_1^2 - 8m_1 - 3 = 0$$(3m_1 + 1)(m_1 - 3) = 0$So,$m_1 = 3$ or $m_1 = -1/3$. Since $m_1 m_2 = -1$,the slopes are $3$ and $-1/3$.The equations of lines passing through $P(2, 1)$ are $y - 1 = 3(x - 2)$ and $y - 1 = -1/3(x - 2)$.$(y - 3x + 5)(3y - 3 + x - 2) = 0$$(y - 3x + 5)(3y + x - 5) = 0$$3y^2 + xy - 5y - 9xy - 3x^2 + 15x + 15y + 5x - 25 = 0$$-3x^2 + 3y^2 - 8xy + 20x + 10y - 25 = 0$Multiplying by $-1$,we get $3x^2 - 3y^2 + 8xy - 20x - 10y + 25 = 0$.

Let $PQR$ be an isosceles right-angled triangle with a 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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