Verified Solution • 2025–26 Session moderate

QJEEM22Ma61e6: JEE Main 2022 Mathematics - Class 11 JEEM22Ma61e6 Solution: Let a triangle be bounded by the lines L 1 : 2x + 5y = 10; L 2…

Let a triangle be bounded by the lines L₁ : 2x + 5y = 10; L₂ : $-$ 4x + 3y = 12 and the line L₃, which passes through the point P(2, 3), intersects L₂ at A and L₁ at B. If the point P divides the line-segment AB, internally in the ratio 1 : 3, then the area of the triangle is equal to :

Multiple Choice Options

A ${{110} \over {13}}$
B ${{132} \over {13}}$
C ${{142} \over {13}}$
D ${{151} \over {13}}$

Step-by-Step Verified Solution

L_{1}: 2 x+5 y=10

L_{2}: -4 x+ 3 y=12

JEE Main 2022 (Online) 28th June Evening Shift Mathematics - Straight Lines and Pair of Straight Lines Question 61 English Explanation

Solving

L_{1}

and

L_{2}

we get

C \equiv\left(\frac{-15}{13}, \frac{32}{13}\right)

Now, Let

A\left(x_{1}, \frac{1}{3}\left(12+4 x_{1}\right)\right)

and

\begin{aligned}&amp;B\left(x_{2}, \frac{1}{5}\left(10-2 x_{2}\right)\right) \\\\&amp;\therefore \quad \frac{3 x_{1}+x_{2}}{4}=2 \\\\&amp;\text { and } \frac{\left(12+4 x_{1}\right)+\frac{10-2 x_{2}}{5}}{4}=3\end{aligned}

So,

3 x_{1}+x_{2}=8

and

10 x_{1}-x_{2}=-5

So,

\left(x_{1}, x_{2}\right)=\left(\frac{3}{13}, \frac{95}{13}\right)

\begin{aligned}&amp;A=\left(\frac{3}{13}, \frac{56}{13}\right) \text { and } B=\left(\frac{95}{13}, \frac{-12}{13}\right) \\\\&amp;=\left|\frac{1}{2}\left(\frac{3}{13}\left(\frac{-44}{13}\right) \frac{-56}{13}\left(\frac{110}{13}\right)+1\left(\frac{2860}{169}\right)\right)\right| \\\\&amp;=\frac{132}{13} \text { sq. units }\end{aligned}

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