1. ⇒ (MHT CET 2023 12th May Morning Shift )

Let $PQR$ be a right angled isosceles triangle, right angled at $P (2, 1)$ . If the equation of the line $QR$ is $2 x + y = 3$ , then the equation representing the pair of lines $P Q$ and $P R$ is

A. $3 x^{2} - 3 y^{2} + 8 x y + 20 x + 10 y + 25 = 0$

B. $3 x^{2} - 3 y^{2} + 8 x y - 20 x - 10 y + 25 = 0$

C. $3 x^{2} - 3 y^{2} + 8 x y + 10 x + 15 y + 20 = 0$

D. $3 x^{2} - 3 y^{2} - 8 x y - 10 x - 15 y - 20 = 0$

Correct answer option is (B)

MHT CET 2023 12th May Morning Shift Mathematics - Straight Lines and Pair of Straight Lines Question 2 English Explanation

Slope of $QR = - 2$ .

Slope of $PQ = m_{1}$

$\begin{aligned} ∴ \tan 45^{\circ} = | \frac{m_{1} + 2}{1 + m_{1} (- 2)} | \\ \Rightarrow 1 = | \frac{m_{1} + 2}{1 - 2 m_{1}} | \\ \Rightarrow m_{1} = - \frac{1}{3} \end{aligned}$

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

$\begin{aligned} y - 1 = - \frac{1}{3} (x - 2) \\ \Rightarrow 3 (y - 1) + (x - 2) = 0 .... (i) \end{aligned}$

Slope of $PR = m_{2} = 3 . . . [∵ P Q ⊥ P R]$

$∴$ equation of $PR$ is

$\begin{aligned} y - 1 = 3 (x - 2) \\ \Rightarrow (y - 1) - 3 (x - 2) = 0 .... (ii) \end{aligned}$

$∴$ The joint equation of the lines is

$\begin{aligned} [3 (y - 1) + (x - 2)] [(y - 1) - 3 (x - 2)] = 0 \\ \Rightarrow 3 (y - 1)^{2} - 8 (y - 1) (x - 2) - 3 (x - 2)^{2} = 0 \\ \Rightarrow 3 (x^{2} - 4 x + 4) + 8 (x y - x - 2 y + 2) \\ - 3 (y^{2} - 2 y + 1) = 0 \\ \Rightarrow 3 x^{2} - 3 y^{2} + 8 x y - 20 x - 10 y + 25 = 0 \end{aligned}$

2. ⇒ (MHT CET 2023 11th May Evening Shift )

The joint equation of the lines pair of lines passing through the point $(3, - 2)$ and perpendicular to the lines $5 x^{2} + 2 x y - 3 y^{2} = 0$ is

A. $3 x^{2} + 2 x y + 5 y^{2} + 14 x + 26 y + 5 = 0$

B. $3 x^{2} + 2 x y - 5 y^{2} - 14 x - 26 y - 5 = 0$

C. $3 x^{2} - 2 x y - 5 y^{2} - 14 x - 26 y + 5 = 0$

D. $3 x^{2} - 2 x y + 5 y^{2} + 14 x + 26 y - 5 = 0$

Correct answer option is (B)

Joint equation of the liens passing through the point $(x_{1}, y_{1})$ and perpendicular to the lines $a x^{2} + 2 h x y + b y^{2} = 0$ is:

$b {(x - x_{1})}^{2} - 2 h (x - x_{1}) (y - y_{1}) + a {(y - y_{1})}^{2} = 0$

$∴$ Equation of the required line is:

$\begin{array}{r} - 3 (x - 3)^{2} - 2 (x - 3) (y + 2) + 5 (y + 2)^{2} = 0 \\ ∴ - 3 (x^{2} - 6 x + 9) - 2 (x y + 2 x - 3 y - 6) + 5 (y^{2} + 4 y + 4) = 0 \end{array}$

$\begin{aligned} ∴ - 3 x^{2} + 18 x - 27 - 2 x y - 4 x + 6 y + 12 + 5 y^{2} \\ + 20 y + 20 = 0 \\ ∴ 3 x^{2} + 2 x y - 5 y^{2} - 14 x - 26 y - 5 = 0 \end{aligned}$

3. ⇒ (MHT CET 2023 10th May Morning Shift )

The number of integral values of $p$ in the domain $[- 5, 5]$ , such that the equation $2 x^{2} + 4 x y - p y^{2} + 4 x + q y + 1 = 0$ represents pair of lines, are

A. 3

B. 4

C. 7

D. 8

Correct answer option is (D)

Given equation of pair of lines is

$2 x^{2} + 4 x y - p y^{2} + 4 x + q y + 1 = 0$

Comparing with

$a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$ , we get $a = 2, h = 2, b = - p$

If the given equation represents a pair of straight lines, then

$\begin{aligned} h^{2} \geq ab \\ \Rightarrow 4 \geq - 2 p \\ \Rightarrow 2 \geq - p \\ \Rightarrow p \geq - 2 \end{aligned}$

$∴$ Possible values of $p$ from domain $[- 5, 5]$ are $- 2, - 1, 0, 1, 2, 3, 4, 5$ .

$∴$ Number of integral values of $p = 8$

⇒Pair of Straight Line