JEE Main 2022 · 28 Jun · Shift 2 · Mathematics Q12

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MathematicsSingle correct+4 / -1Question 12 of 90

Let a triangle be bounded by the lines $L_1:2x+5y=10$ ; $L_2:-4x+3y=12$ and the line $L_3$ , which passes through the point $P(2,3)$ , intersects $L_2$ 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 :

\dfrac{110}{13}

\dfrac{132}{13}

\dfrac{142}{13}

\dfrac{151}{13}

Answer and worked solution

Official answer: B

Points. $A=\left(a,\frac{12+4a}3\right)$ on $L_2$ , $B=\left(b,\frac{10-2b}5\right)$ on $L_1$ , and $P=\frac{3A+B}4$ : $3a+b=8$ and $10a-b=-5$ , so $a=\frac3{13}$ , $b=\frac{95}{13}$ .

Vertices (in units of $\frac1{13}$ ): $A(3,56)$ , $B(95,-12)$ , and $L_1\cap L_2=(-15,32)$ .

Area. $\frac12\cdot\frac{|92(-24)-(-68)(-18)|}{169}=\frac{1716}{169}=\frac{132}{13}$ .

(B)

Topics:Distance and section formulae Equations of a line

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JEE Main 2022 · 28 Jun · Shift 2, Mathematics question 12