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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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 $m_1 m_2 = -1$. Let $m_1 = m$,then $m_2 = -\frac{1}{m}$.Since $	riangle PQR$ is isosceles with $\angle P = 90^\circ$,we have $\angle PQR = \angle PRQ = 45^\circ$.The slope of line $QR$ is $m_{QR} = -2$.The angle between $PQ$ (slope $m$) and $QR$ (slope $-2$) is $45^\circ$:$	an 45^\circ = \left| \frac{m - (-2)}{1 + m(-2)} ight| = 1$$\left| \frac{m+2}{1-2m} ight| = 1 \Rightarrow (m+2)^2 = (1-2m)^2$$m^2 + 4m + 4 = 1 - 4m + 4m^2 \Rightarrow 3m^2 - 8m - 3 = 0$$(3m+1)(m-3) = 0 \Rightarrow m = 3, -\frac{1}{3}$.The equations of lines $PQ$ and $PR$ passing through $P(2, 1)$ are $y-1 = 3(x-2)$ and $y-1 = -\frac{1}{3}(x-2)$.These can be written as $3x - y - 5 = 0$ and $x + 3y - 5 = 0$.The joint equation is $(3x - y - 5)(x + 3y - 5) = 0$.$3x^2 + 9xy - 15x - xy - 3y^2 + 5y - 5x - 15y + 25 = 0$$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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