Let a triangle be bounded by the lines $\mathrm{L}_{1}: 2 \mathrm{x}+5 \mathrm{y}=10$; $\mathrm{L}_{2}:-4 \mathrm{x}+3 \mathrm{y}=12$ and the line $\mathrm{L}_{3}$, which passes through the point $\mathrm{P}(2,3)$, intersect $\mathrm{L}_{2}$ at A and $\mathrm{L}_{1}$ 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

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Straight Lines

Let a triangle be bounded by the lines

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

;

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

and the line

L_{3}

, which passes through the point

P (2, 3)

, intersect

L_{2}

at A and

L_{1}

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

Answer: (B)

Point A lies on

L_{2}

A (α, 4 + \frac{4}{3} α)

Point B lies on

L_{1}

B (β, 2 - \frac{2}{5} β)

Since P(2,3) divides AB in ratio 1:3:

P (2, 3) = (\frac{3 α + β}{4}, \frac{3 ( 4 + \frac{4}{3} α ) + ( 2 - \frac{2}{5} β )}{4})

Solving:

α = \frac{3}{13}

β = \frac{95}{13}

A (\frac{3}{13}, \frac{56}{13})

B (\frac{95}{13}, - \frac{12}{13})

Vertex C is the intersection of

L_{1}

and

L_{2}

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

Area of triangle ABC:

Area = \frac{1}{2} \frac{3}{13} \frac{95}{13} - \frac{15}{13} \frac{56}{13} - \frac{12}{13} \frac{32}{13} 111 = \frac{132}{13}

sq. units.

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Single correct option