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

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Solution

The formula for the area ‘A’ encompassed by three points, and is given by the formula,
$∆ = \frac{1}{2} |(x_{1} y_{2} + x_{2} y_{3} + x_{3} y_{1}) - (x_{2} y_{1} + x_{3} y_{2} + x_{1} y_{3})|$

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

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

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