MHT-CET Full Test-7 Mathematics Que-50 Solution

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

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

C. 3x^2 - 3y^2 + 8xy + 10x + 15y + 20 = 0

D. 3x^2 - 3y^2 - 8xy - 10x - 15y - 20 = 0

Answer :- B. 3x^2 - 3y^2 + 8xy - 20x - 10y + 25 = 0

Explanation :-

Slopeof QR = -2

Slope of PQ = m₁

\tan{45^\circ} = \frac{m_1 + 2}{|1 + m_1(-2)|} \Rightarrow 1 = \frac{m_1 + 2}{1 - 2m_1} \Rightarrow m_1 = -\frac{1}{3}

Equation of PQ passing through point P(2, 1) and having slope m_1 = -\frac{1}{3} is

y - 1 = -\left(x - 2\right) \Rightarrow 3(y - 1) + (x - 2) = 0 \dots \text{(i)}

Slope of PR = m_2 = 3 … \text{[PQ} \perp \text{ PR]}

Equation of PR is

y - 1 = 3(x - 2) \Rightarrow (y - 1) - 3(x - 2) = 0 \dots \text{(ii)}

The joint equation of the lines is

\left[(3(y - 1) + (x - 2))\right] \left[(y - 1) - 3(x - 2)\right] = 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

3x² – 3y² + 8xy-20x-10y +25 = 0