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*.

#### Solution

GIVEN: The four vertices of quadrilateral are (1, 2), (−5, 6), (7, −4) and D (*k*, −2) taken in order. If the area of the quadrilateral is zero

TO FIND: value of *k*

PROOF: Let four vertices of quadrilateral are A (1, 2) and B (−5, 6) and C (7, −4) and D (*k*, −2)

We know area of triangle formed by three points (x_{1}, y_{1}),(x_{2}, y_{2}) and (x_{3}, y_{3})is given by

`Δ =1/2[x_1(y_2 -y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)`]

Δ`=1/2[x_1y_2+x_2y_3+x_3y_3)]`

NOW AREA OF ΔABC

Taking three point when A(1,2) and B (-5,6) and C (7,-4)

Area (ΔABC)

`=1/2[{6+20+14}-{-10+42-4}`

`=1/2[{40}-{28}]`

`=1/2[{12}]`

Area (ΔABC) =6 sq.units

Also,

Now Area of ΔACD

Taking three points when A (1, 2) and C (7, −4) and D (*k*, −2)

`=1/2[{-4-14+2k}-{14-4k-2}`

`=1/2[{2k-18}-{12-4k}`

`=1/2[{6k-30}]`

`=[{3k-15}]`

Hence

Area (ABCD) = Area (ΔABC) + Area (ΔACD)

`0=6+3k-15` (substituting the value )

`k=3`