Question

Find the value of k for which the points (3,2), (1,2), (5,k) are collinear.

(a) 2

(b) 5

(c) 4

(d) 9

I had been asked this question by my college professor while I was bunking the class.

Enquiry is from Area of a Triangle in chapter Determinants of Mathematics – Class 12

Answer 1

The correct option is (a) 2

For explanation I would say: Given that the vertices are (3,2), (1,2), (5,k)

Therefore, the area of the triangle with vertices (3,2), (1,2), (5,k) is given by

Δ=\(\frac{1}{2}\) \(\begin{Vmatrix}3&2&1\\1&2&1\\5&k&1\end{Vmatrix}\)=0

Applying R1→R1-R2, we get

\(\frac{1}{2}\begin{Vmatrix}2&0&0\\1&2&1\\5&k&1\end{Vmatrix}\)=0

Expanding along R1, we get

\(\frac{1}{2}\) {2(2-k)-0+0}=0

2-k=0

k=2.