Find the value of k whose (7,-2),(5,1)and (3,k) ate collinear
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Since the given points are in collinear then the area of triangle is zero
Area of triangle = 1/2 [x1(y2-y3) +x2(y3-y1) +x3(y1-y2)]
0 = 1/2[7(1-k) +5(k +2) +3(-2 - 1)]
0 = 7(1-k) +5(k+2) +3(-3)
0 = 7-7k +5k +20 - 9
0 = 7+20 - 9 - 7k +5k
0 = 18 - 3k
- 18 = - 3k
- 18/-3 = k
6 = k
Therefore the value of k = 6
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