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Question

The value of k for which the points A(2,3),B(4,k), and C(6,3) are collinear.

A

k = 9

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B

k = -6

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C

k = 0

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D

k = 4

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Solution

The correct option is C: k=0

Given three points are A(2,3),B(4,k), and C(6,3).

Here, x1=2,y1=3,x2=4,y2=k,x3=6,y3=3

When the points are collinear, the area of the triangle will be zero.

Area of the triangle =12[x1(y2y1)+x2(y3y1)+x3(y1y2)]

The points are collinear.

12[x1(y2y1)+x2(y3y1)+x3(y1y2)]=0

12[2(k3)+2(33)+6(3k)]=0

12[2k612+186k]=0

|2k|=0 [Area cannot be negative]

2k=0

k=0


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