PQR is a right-angled isosceles triangle with | vedclass

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(C) Let the slope of $PQ$ be $m$. Since $PQ \perp PR$ and $PQR$ is isosceles,the slope of $PR$ is $-1/m$. The angle between $PQ$ and $QR$ ($2x + y = 3

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let the slope of $PQ$ be $m$. Since $PQ \perp PR$ and $PQR$ is isosceles,the slope of $PR$ is $-1/m$. The angle between $PQ$ and $QR$ ($2x + y = 3$,slope $-2$) is $45^\circ$. Using $	an 	heta = |(m_1 - m_2) / (1 + m_1 m_2)|$,we have $	an 45^\circ = |(m - (-2)) / (1 + m(-2))| = 1$. This gives $|(m + 2) / (1 - 2m)| = 1$. Solving $m + 2 = 1 - 2m$ gives $3m = -1 \Rightarrow m = -1/3$. Solving $m + 2 = -(1 - 2m)$ gives $m + 2 = -1 + 2m \Rightarrow m = 3$. The lines $PQ$ and $PR$ pass through $P(2, 1)$ with slopes $3$ and $-1/3$. Their equations are $(y - 1) = 3(x - 2) \Rightarrow 3x - y - 5 = 0$ and $(y - 1) = -1/3(x - 2) \Rightarrow x + 3y - 5 = 0$. The joint 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 $3x^2 - 3y^2 + 8xy - 20x - 10y + 25 = 0$.

$PQR$ is 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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