Question

If A(–7, 5), В(–6, –7), С(–3, –8) and D(2, 3) are the vertices of a quadrilateral ABCD, then find the area of the quadrilateral.

Answer 1

Given: A(–7, 5), В(–6, –7), С(–3, –8) and D(2, 3)

Step-wise calculation:

1. Use the shoelace formula for vertices in order A→B→C→D→A:

Area = `1/2 |(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)|`

2. Compute the two sums:

x₁y₂ = (–7)(–7)

= 49 

x₂y₃ = (–6)(–8)

= 48 

x₃y₄ = (–3)(3)

= –9 

x₄y₁ = (2)(5)

= 10 

Sum1 = 49 + 48 – 9 + 10

= 98

y₁x₂ = (5)(–6)

= –30 

y₂x₃ = (–7)(–3)

= 21 

y₃x₄ = (–8)(2)

= –16 

y₄x₁ = (3)(–7)

= –21 

Sum2 = –30 + 21 – 16 – 21

= –46

3. Area = `1/2` |Sum1 – Sum2| 

= `1/2 |98 - (-46)|`

= `1/2 |144|`

= 72

The area of quadrilateral ABCD is 72 square units.