The four vertices of a quadrilateral are (1, 2), (−5, 6), (7, −4) and (k, −2) taken in order.

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 $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ is given by

$\Delta=\frac{1}{2}\left|x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right|$

$\Delta=\frac{1}{2}\left|\left(x_{1} y_{2}+x_{2} y_{3}+x_{3} y_{1}\right)-\left(x_{1} y_{3}+x_{2} y_{1}+x_{3} y_{2}\right)\right|$

Now Area of ΔABC

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

Area $(\triangle A B C)$

$=\frac{1}{2}\{6+20+14\}-\{-10+42-4\} \mid$

$=\frac{I}{2}\{40\}-\{28\} \mid$

$=\frac{1}{2}\{12\} \mid$

Area $(\triangle A B C)=6$ sq. units

Also,

Now Area of ΔACD

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

$\operatorname{Arca}(\Delta A C D)$

$=\frac{1}{2}\{-4-14+2 k\}-\{14-4 k-2\} \mid$

$=\frac{1}{2}\{2 k-18\}-\{12-4 k\} \mid$

$=\frac{I}{2}\{6 k-30\} \mid$

$=\{3 k-15\} \mid$

Hence

Area $(A B C D)=\operatorname{Arca}(\triangle A B C)+\operatorname{Area}(\triangle A C D)$

$0=6+3 k-15$ (substituting the values)

$k=3$