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

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(B) Since $	riangle PQR$ is an isosceles right-angled triangle with the right angle at $P$,we have $\angle PQR = \angle PRQ = 45^{\circ}$.Let $m$ be

Vedclass · Accepted Answer

$3x - y - 5 = 0$

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(B) Since $	riangle PQR$ is an isosceles right-angled triangle with the right angle at $P$,we have $\angle PQR = \angle PRQ = 45^{\circ}$.Let $m$ be the slope of the line $QR$. Given $2x + y = 3$,we have $y = -2x + 3$,so $m = -2$.Let $m_1$ be the slope of the line $PQ$. The angle between $PQ$ and $QR$ is $45^{\circ}$.Using the formula $	an 	heta = |\frac{m_1 - m}{1 + m_1 m}|$,we have:$	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}|$This gives two cases:Case $1$: $\frac{m_1 + 2}{1 - 2m_1} = 1$ $\Rightarrow m_1 + 2 = 1 - 2m_1$ $\Rightarrow 3m_1 = -1$ $\Rightarrow m_1 = -\frac{1}{3}$.Case $2$: $\frac{m_1 + 2}{1 - 2m_1} = -1$ $\Rightarrow m_1 + 2 = -1 + 2m_1$ $\Rightarrow m_1 = 3$.Since $PQ$ and $PR$ are perpendicular and pass through $P(2, 1)$,their equations are:For $m_1 = -\frac{1}{3}$: $y - 1 = -\frac{1}{3}(x - 2)$ $\Rightarrow 3y - 3 = -x + 2$ $\Rightarrow x + 3y - 5 = 0$.For $m_2 = 3$: $y - 1 = 3(x - 2)$ $\Rightarrow y - 1 = 3x - 6$ $\Rightarrow 3x - y - 5 = 0$.Comparing with the options,$3x - y - 5 = 0$ is the correct equation.

Let $PQR$ be a right-angled isosceles triangle,right-angled at $P(2, 1)$. If the equation of the side $QR$ is $2x + y = 3$,then the equation of one of the sides other than $QR$ is

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