Question

find the area of quadrilateral whose vortices are A(-3,1) B(-2,-2) C(1,4) D(3,-1) ​

Answer 1

Solution:-

Let the Vertices of Quadrilateral be A( -3 ,1) ,B(-2,-2) ,C (1 ,4) & D(3 ,-1)

Joining AC

There are Two Triangle Formed ∆ABC &∆ACD

• Hence Ar of Quadrilateral= ar(∆ABC) + ar(∆ACD)

Finding ar(∆ABC)

Area of Triangle ABC = 1/2 [x1(y2 - y1) + x2 (y3 - y1) +x3( y1 - y2)

Here x1 = -3 , x2 = -2 & x3 = 1

y1 = 1 , y2 = -2 & y3 = 4

Putting Values:-

= 1/2 [ -3(-2 - 1)+ (-2)(4 - 1) + 1(1 - (-2)

= 1/2 [ -3 × 3 + (-2) ×3 + 3]

= 1/2 [ -9 - 6 + 3]

= 1//2 [ -9 -3]

= 1/2 ×12

= 6 square units

Finding ar(∆ADC)

Here x1 = -3 , x2 = 3 & x3 = 1

y1 = 1 , y2 = -1 & y3 = 4

Putting Values:-

= 1/2 [ -3(-1 -1) + 3(4 - 1)+1(1 - (-1)

= 1/2 [ -3×2 + 3 × 3 + 2]

= 1/2 [ -6 + 9 + 2]

= 1/2 [ -6 + 11]

= 1/2 (5)

Area of Quadrilateral = ar(∆ABC) + (ar∆ADC)

=> 5/2 + 6

=> 15 square units

Area of Quadrilateral is 15 square units