Question

Find the area of the triangle formed by the mid points of sides of the triangle whose vertices are
(2, 1), (-2, 3), (4, -3)
(A) 1.5 sq. units
(B) 3'sq. units
(C) 6 sq. units
(D) 12 sq. units​

Answer 1

Answer:

Area of the triangle is 6 square units.

Step-by-step explanation:

Given coordinates are :-

A(2,1), B(-2,3) and C(4,-3)

Formula for the area of triangle :$= > \frac{1}{2}[x_{1}(y_{2} - y_{3}) + x_{2}(y_{3} - y_{1}) + x_{3}(y_{1} - y_{2})]$

Here (x1, y1), (x2, y2) & (x3, y3) are the coordinates of triangle.

Therefore:-

(x1, y1) is (2,1)

(x2,y2) is (-2,3)

(x3,y3) is (4,-3)

So on using the formula:-

$= > \frac{1}{2}[x_{1}(y_{2} - y_{3}) + x_{2}(y_{3} - y_{1}) + x_{3}(y_{1} - y_{2})]$

=> 1/2[2(3 + 3)+{-2(-3-1)} + 4(1 - 3]

=> 1/2[2(6)- 2(-4) + 4(-2)]

=> 1/2[12 + 8 - 8]

=> 1/2[12]

=> 12/2

=> 6 square units.

Therefore:-

Area of the triangle is 6 square units.