Let the points of intersections of the lines x - y + 1 = 0, x - 2y + 3 = 0 and 2x - 5y + 11 = 0 are the mid points of the sides of a triangle \Delta ABC. Then, the area of the \Delta ABC is \underline\hspace2cm .

Straight Lines and Pair of Straight Lines PYQ — JEE Main 2021 · 1 Sep · Shift 2 · Q41 with Solution

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Correct answer: 6

The given lines are: $L_1: x-y+1=0$ $L_2: x-2y+3=0$ $L_3: 2x-5y+11=0$

These pairwise intersections are the midpoints of the sides of triangle $ABC$ .

Let

From $L_1$ : $x-y+1=0 \implies x=y-1$

Substitute into $L_2$ : $x-2y+3=0$ $y-1-2y+3=0$ $-y+2=0 \implies y=2$ So, $x=2-1=1$ Hence, $P=(1,2)$

From $L_2$ : $x-2y+3=0 \implies x=2y-3$

Substitute into $L_3$ : $2x-5y+11=0$ $2(2y-3)-5y+11=0$ $4y-6-5y+11=0$ $-y+5=0 \implies y=5$ So, $x=2(5)-3=7$ Hence, $Q=(7,5)$

From $L_1$ : $x=y-1$

Substitute into $L_3$ : $2x-5y+11=0$ $2(y-1)-5y+11=0$ $2y-2-5y+11=0$ $-3y+9=0 \implies y=3$ So, $x=3-1=2$ Hence, $R=(2,3)$

The points $P,Q,R$ are the midpoints of the sides of triangle $ABC$ . Therefore, triangle $PQR$ is the medial triangle of $ABC$ .

A medial triangle has area equal to one-fourth of the original triangle: $[PQR]=\frac14[ABC]$ So, $[ABC]=4[PQR]$

Using coordinates: $P(1,2),\ Q(7,5),\ R(2,3)$

Area formula: $[PQR]=\frac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$

Substitute: $[PQR]=\frac12\left|1(5-3)+7(3-2)+2(2-5)\right|$ $=\frac12\left|2+7-6\right|$ $=\frac12\cdot 3=\frac32$

$[ABC]=4\times \frac32=6$

So, the required area is: $\boxed{6}$