Find the value of k, if the points A (8, 1), B (3, -4) and C (2, k) are collinear.

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Solution

If 3 points are collinear, then the area of the triangle formed by them is zero.

Area of triangle = $\frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) | = 0$

$\Rightarrow x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 0$

$\Rightarrow 8 (- 4 - k) + 3 (k - 1) + 2 (1 + 4) = 0$

$\Rightarrow 5 k = - 25$

$∴ k = - 5$

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Similar questions

A (8, 1)

B (3, - 2 k)

C (k, - 5)

Show that the following points are collinear :

(i) A (2, -2), B(-3, 8) and C(-1, 4)

(ii) A(-5, 1), B(5,5) and C(10, 7)

(iii) A(5, 1), B(1, -1) and C(11, 4)

(iv) A(8, 1), B(3, -4) and C(2, -5)

K

(- 3, 11)

(6, 2)

(K, 4)

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