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Find the value of ‘k’ for which the points A(–1, 3), B(2, k) and C(5, –1) are collinear. |
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Answer» Given that, Points A(–1, 3), B(2, k) and C(5, –1) are collinear. i.e., they will not form any triangle . ∴ Area of triangle ABC = 0 ∴ \(\frac{1}{2}\) [−1(k − (−1)) + 2(−1 − 3) + 5(3 − k))] = 0 (By formula of area of triangle passing through 3 vertices) ⇒ −k − 1 − 8 + 15 − 5k = 0 ⇒ −6k + 6 = 0 ⇒ k = 1. Hence, The value of k = 1, for which given points A, B & C are collinear. |
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