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Question

Find the values of k, if the points A(k+1, 2k), B(3k,2k+3) and C(5k-1,5k) are collinear.

A
2
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B
12
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C
4
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D
14
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Solution

The correct options are
A 2
B 12


We know that, if three points are collinear, then the area of the triangle formed by these points is zero.
Since, the points A(k+1,2k), B(3k,2k+3) and C(5k-1,5k), are collinear.
Then, area of Δ ABC=0.
12[x1(y2y3)+x2(y3y1)+x3(y1y2)]=0Here, x1=k+1,x2=3k,x3=5k1 and y12k1,y22k+3,y3=5k12[(k+1)(2k+35k)+3k(5k2k)+(5k1)(2k(2k+3))]=012[(k+1)(33k)+3k(3k)+(5k1)(2k2k3)]=012[3k2+3k3k+3+9k215k+3]=012(6k215k+6)=0 [multiply by 2]6k215k+6=0 [by factorization method]2k25k+2=02k24kk+2=02k(k2)1(k2)=0(k2)(2k1)=0
If k – 2 = 0, then k = 2
If 2k – 1, then k = 12
k=2,12
Hence, the required values of k are 2 and 12.

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