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$$\begin{aligned} & 2=\frac{2 x_2+x_1}{3}, \frac{-4}{3}=\frac{2 y_2+y_1}{3} \\ & 2 x_2+x_1=6,2 y_2+y_1=-4 \end{aligned}$$

$$\begin{aligned} & x_1=6-2 x_2 \quad \text{.... (1)}\\ & y_1=-4-2 y_2 \quad \text{.... (2)}\\ & 4 x_1+y_1=14 \quad \text{.... (3)}\\ & 3 x_2-2 y_2=5 \quad \text{.... (4)} \end{aligned}$$

From here, $$x_2=1, y_2=-1, x_1=4, y_1=-2$$

$$\begin{aligned} & B(4,-2) C(1,-1) \\ & y+2=\frac{-1+2}{1-4}(x-4) \\ & -3 y-6=x-4 \\ & x+3 y+2=0 \end{aligned}$$

19:T41a,

JEE Main 2024 (Online) 8th April Morning Shift Mathematics - Straight Lines and Pair of Straight Lines Question 36 Hindi Explanation

$$\begin{aligned} & 2=\frac{2 x_2+x_1}{3}, \frac{-4}{3}=\frac{2 y_2+y_1}{3} \\ & 2 x_2+x_1=6,2 y_2+y_1=-4 \end{aligned}$$

$$\begin{aligned} & x_1=6-2 x_2 \quad \text{.... (1)}\\ & y_1=-4-2 y_2 \quad \text{.... (2)}\\ & 4 x_1+y_1=14 \quad \text{.... (3)}\\ & 3 x_2-2 y_2=5 \quad \text{.... (4)} \end{aligned}$$

यहाँ से, $$x_2=1, y_2=-1, x_1=4, y_1=-2$$

$$\begin{aligned} & B(4,-2) C(1,-1) \\ & y+2=\frac{-1+2}{1-4}(x-4) \\ & -3 y-6=x-4 \\ & x+3 y+2=0 \end{aligned}$$

Straight Lines And Pair Of Straight Lines

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