Q51 / 90 · JEE Main 2021 (01 Sep Shift 2) · Mathematics

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

Correct answer: 6

Solution

If $P, Q, R$ are mid-points of sides of $∆ A B C$ then $4 \times A r e a (ΔPQR) = A r e a (∆ ABC)$

Area of $Δ ABC = 4 \times \frac{1}{2} |\frac{{|\begin{array}{lll} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array}|}^{2}}{C_{1} C_{2} C_{3}}|$

Where $C_{1}, C_{2}, C_{3}$ are cofactors of $c_{1}, c_{2}, c_{3}$

$Δ ABC = 4 \times \frac{1}{2} |\frac{{|\begin{array}{lll} 1 & - 1 & 1 \\ 1 & - 2 & 3 \\ 2 & - 5 & 11 \end{array}|}^{2}}{(- 1) (3) (- 1)}|$

$= 2 \times 9 \div 3$

$= 6$

Hence, the correct answer is 6.

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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 ABC. Then the area of the triangle ABC is

Solution

Relevant Questions

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