1.

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

Answer»

Given points are A(8,1),B(3,−4) and C(2,k).It is also said that they are collinear and hence the area enclosed by them should be 0. 

Area of the triangle having vertices (x1,y1), (x2,y2) and (x3,y3

\(\frac{1}2\) |x1(y2-y3)+x2(y3-y1)+x3(y1-y2)| 

Given that area of ∆ABC = 0 

∴ 0 = \(\frac{1}2\) |8(-4 – k) + 3(k – 1) + 2(1 – (-4))| 

∴ 0 = \(\frac{1}2\) |-32 – 8k + 3k -3 + 10| 

∴ 5k + 25 = 0 

∴ k = -5 

Hence, 

the value of k is -5.



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