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

Question

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  (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

intersection point of give lines are (1,2),(7,5) (2,3) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/bfa97fba9850ee9065bc689db5eef334-.png" /> Δ=(1/2)|1 2 1 7 5 1 2 3 1| =(1/2)[1(5-3)-2(7-2)+1(21-10)] =(1/2)[2-10+11] Δ DEF =(1/2)(3)=(3/2) Δ ABC =4 Δ DEF =4((3/2))=6

Q. 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 $A BC$ . Then the area of the triangle $A BC$ is ___

intersection point of give lines are $(1, 2), (7, 5)$ $(2, 3)$

$Δ = \frac{1}{2} ∣ ∣ 172253111 ∣ ∣$
$= \frac{1}{2} [1 (5 - 3) - 2 (7 - 2) + 1 (21 - 10)]$
$= \frac{1}{2} [2 - 10 + 11]$
$Δ D EF = \frac{1}{2} (3) = \frac{3}{2}$
$Δ A BC = 4Δ D EF = 4 (\frac{3}{2}) = 6$

Q. 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 ___

intersection point of give lines are (1,2),(7,5) (2,3) Δ=21​∣∣​172​253​111​∣∣​ =21​[1(5−3)−2(7−2)+1(21−10)] =21​[2−10+11] ΔDEF=21​(3)=23​ ΔABC=4ΔDEF=4(23​)=6

Q. 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 $A BC$ . Then the area of the triangle $A BC$ is ___

intersection point of give lines are $(1, 2), (7, 5)$ $(2, 3)$

$Δ = \frac{1}{2} ∣ ∣ 172253111 ∣ ∣$
$= \frac{1}{2} [1 (5 - 3) - 2 (7 - 2) + 1 (21 - 10)]$
$= \frac{1}{2} [2 - 10 + 11]$
$Δ D EF = \frac{1}{2} (3) = \frac{3}{2}$
$Δ A BC = 4Δ D EF = 4 (\frac{3}{2}) = 6$