Find the value of k if points A(k, 3), B(6, −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(k, 3), B(6, −2) and C(−3, 4). It is also said that they are collinear and hence the area enclosed by them should be 0.

$\Delta=\frac{1}{2}|(k(-2)+6 \times 4+(-3) \times 3)-(6 \times 3+(-3)(-2)+k \times 4)|$

$0=\frac{1}{2}|(-2 k+24-9)-(18+6+4 k)|$

$0=\frac{1}{2}|-2 k+15-24-4 k|$

$0=\frac{1}{2}|-6 k-9|$

$0=-6 k-9$

$k=-\frac{9}{6}=-\frac{3}{2}$

Hence the value of ' $k$ ' for which the given points are collinear is $k=-\frac{3}{2}$.