Find the value of k, if the points A (8, 1) B(3, −4)

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(8,1), B(3,−4) and C(2,k). It is also said that they are collinear and hence the area enclosed by them should be 0.

$\Delta=\frac{1}{2}|(8 \times-4+3 \times k+2 \times 1)-(3 \times 1+2 \times-4+8 \times k)|$

$0=\frac{1}{2}|(-32+3 k+2)-(3-8+8 k)|$

$0=\frac{1}{2}|-25-5 k|$

$k=-5$

Hence the value of ' $k$ ' for which the given points are collinear is $k=-5$.