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

Let A (1, 2) , B (-5, 6), C (7, -4) and D (k , -2) be the vertices of the quadrilateral.

We have,

$\therefore$ Area of $△$ ABC =$\frac{1}{2}\left[\right\{(6+20+14)-(-10+42-4)\left\}\right]$
$\Rightarrow Areaof△ABC=\frac{1}{2}(40-28)=6sq.units$

Also, we have

Area of $△$ ACD = $\frac{1}{2}\left\{\right(-4-14+2k)-(14-4k-2\left)\right\}$

$\Rightarrow Areaof△ACD=\frac{1}{2}\left\{\right(2k-18)-(12-4k\left)\right\}\Rightarrow Areaof△ACD=\frac{1}{2}(6k-30)=(3k-15)$

$\therefore$ Area of quadrilateral ABCD = 6+ 3k -15 = 3k- 9

It is given that the area of quadrilateral is zero.

$\therefore$ 3k -9 = 0

$\Rightarrow$ k = 3