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

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(B) The slope of line $QR$ is $m = -2$. Let the slopes of lines $PQ$ and $PR$ be $m_1$ and $m_2$ respectively.Since $	riangle PQR$ is an isosceles ri

Vedclass · Accepted Answer

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

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(B) The slope of line $QR$ is $m = -2$. Let the slopes of lines $PQ$ and $PR$ be $m_1$ and $m_2$ respectively.Since $	riangle PQR$ is an isosceles right-angled triangle,the angles $\angle PQR = \angle PRQ = 45^{\circ}$.Using the formula $	an 	heta = \left| \frac{m - m_1}{1 + m \cdot m_1} ight|$,we have $	an 45^{\circ} = \left| \frac{-2 - m_1}{1 + (-2)m_1} ight| = 1$.This gives $\left| \frac{2 + m_1}{1 - 2m_1} ight| = 1$,so $2 + m_1 = 1 - 2m_1$ or $2 + m_1 = -(1 - 2m_1)$.Solving $3m_1 = -1$ gives $m_1 = -1/3$. Solving $-m_1 = -3$ gives $m_2 = 3$.The lines $PQ$ and $PR$ pass through $P(2, 1)$ with slopes $-1/3$ and $3$.The equations are $y - 1 = -1/3(x - 2) \Rightarrow x + 3y - 5 = 0$ and $y - 1 = 3(x - 2) \Rightarrow 3x - y - 5 = 0$.The joint equation is $(x + 3y - 5)(3x - y - 5) = 0$.Expanding this: $3x^2 - xy - 5x + 9xy - 3y^2 - 15y - 15x + 5y + 25 = 0$.Simplifying gives $3x^2 - 3y^2 + 8xy - 20x - 10y + 25 = 0$.

Let $PQR$ be a right-angled isosceles triangle,right-angled 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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