Question

Find the value of k if points A(k, 3), B(6, −2) and C(−3, 4) are collinear.

Answer 1

The formula for the area ‘A’ encompassed by three points, and is given by the formula,
$∆=\frac{1}{2}\left|\left(x_1{y}_2+x_2{y}_3+x_3{y}_1\right)-\left(x_2{y}_1+x_3{y}_2+x_1{y}_3\right)\right|$

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.
$$\begin{gathered} ∆=\frac{1}{2}\left|\left(k\left(-2\right)+6\times 4+\left(-3\right)\times 3\right)-\left(6\times 3+\left(-3\right)\left(-2\right)+k\times 4\right)\right| \\ 0=\frac{1}{2}\left|\left(-2k+24-9\right)-\left(18+6+4k\right)\right| \\ 0=\frac{1}{2}\left|-2k+15-24-4k\right| \\ 0=\frac{1}{2}\left|-6k-9\right| \\ 0=-6k-9 \\ k=-\frac{9}{6}=-\frac{3}{2} \end{gathered}$$

Hence the value of ‘k’ for which the given points are collinear is.