Find the value of $^{'} K^{'}$ for which the points are collinear.
(i) $(7, - 2) (5, 1) (3, K)$

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Solution

$\begin{matrix} A (7, - 2) b (5, 1) C (3, k) m_{A B} = m_{A C} \frac{1 + 2}{5 - 7} = \frac{k + 2}{3 - 7} \frac{- 3}{2} = \frac{- (k + 2)}{4} 6 = k + 2 ∴ k = 4. \end{matrix}$

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k

(7, - 2), (5, 1), (3, k)

In each of the following find the value of ‘k’, for which the points are collinear.

(i) (7, − 2), (5, 1), (3, − k) (ii) (8, 1), (k, − 4), (2, − 5)

k

(4 k - 1, k - 2), (k, k - 7)

(k - 1, - k - 2)

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(7, - 2), (5, 1)

(3, K)

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