Question

The four vertices of a quadrilateral are $(1,2),(-5,6),(7,-4)$ and $(k,-2)$ taken in order. If the area of the quadrilateral is zero, find the value of $k$.

Answer 1

Formula:

Area of quadrilateral = area of triangle 1 + area of triangle 2

Area of triangle $=\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$

Given,

$A(1,2),B(-5,6),C(7,-4),D(k,-2)$

Let the first triangle be, ABC

Area of $△ABC=\frac{1}{2}\left[1\right(6-(-4))-5(-4-2)+7(2-6\left)\right]$

$=\frac{1}{2}[10+30-28]$

$=6$sq.units

Area of $△ACD=\frac{1}{2}\left[1\right(-4+2)+7(-2-2)+k(2+4\left)\right]$

$=\frac{1}{2}[-2-28+6k]$

$=-15+3k$sq.units

Area of quadrilateral $=6-15+3k=0$

$k=3$