Q.50 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
A. 3x2 – 3y2 + 8xy + 20x + 10y + 25 = 0
B. 3x2 – 3y2 + 8xy – 20x – 10y + 25 = 0
C. 3x2 – 3y2 + 8xy + 10x + 15y + 20 = 0
D. 3x2 – 3y2 – 8xy – 10x – 15y – 20 = 0
Answer: B
Explanation:
Slope of \mathrm{QR} = -2.
Slope of \mathrm{PQ} = m_1
\therefore \quad \tan 45^\circ = \left| \frac{m_1 + 2}{1 + m_1(-2)} \right| \Rightarrow 1 = \left| \frac{m_1 + 2}{1 - 2 m_1} \right| \Rightarrow m_1 = -\frac{1}{3} \therefore \quad \text{Equation of PQ passing through point } \mathrm{P}(2, 1) \text{ and having slope } \frac{-1}{3} \text{ is} y-1=-\frac13(x-2) \Rightarrow3(y-1)+(x-2)=0...(i)Slope of \mathrm{PR} = m_2 = 3 \quad \ldots PQ \perp PR
\therefore \quad \text{Equation of PR is } y - 1 = 3(x - 2) \Rightarrow (y - 1) - 3(x - 2) = 0 \therefore \quad \text{The joint equation of the lines is} [3(y - 1) + (x - 2)][(y - 1) - 3(x - 2)] = 0 \Rightarrow 3(y - 1)^2 - 8(y - 1)(x - 2) - 3(x - 2)^2 = 0 \Rightarrow 3(x^2 - 4x + 4) + 8(xy - x - 2y + 2) - 3(y^2 - 2y + 1) = 0 \Rightarrow 3x^2 - 3y^2 + 8xy - 20x - 10y + 25 = 0