Find the value of k, if the points A(7, −2),

The formula for the area ' $A$ ' encompassed 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 the formula,

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

If three points are collinear the area encompassed by them is equal to 0.

The three given points are A(7, −2), B(5, 1) and C(3, 2k). It is also said that they are collinear and hence the area enclosed by them should be 0.

$\Delta=\frac{1}{2}|(7 \times 1+5 \times 2 k+3 \times-2)-(5 \times-2+3 \times 1+7 \times 2 k)|$

$0=\frac{1}{2}|(7+10 k-6)-(-10+3+14 k)|$

$0=\frac{1}{2}|-4 k+8|$

$0=-4 k+8$

$k=2$

Hence the value of ' $k$ ' for which the given points are collinear is $k=2$.