Question

Find the value of $k$ for which the points $(2,3),(3,k)$ and $(3,7)$ are collinear.

• A. 5
• B. 6
• C. 8
• D. 7

Answer 1

Answer: (D)

7

If three points are collinear, then the area of the triangle is zero.

Area of the triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$ and $(x_3,y_3)$ is:

$A=\frac{1}{2}\left|x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right|$

Here the points are $(2,3)$, $(3,k)$ and $(3,7)$ and it is given that these points are collinear therefore the area is zero that is:

$0=\frac{1}{2}\left|2\right(k-7)+3(7-3)+3(3-k\left)\right|\Rightarrow 0=|2k-14+3(4)+9-3k|\Rightarrow 0=|-k-5+12|\Rightarrow 0=|-k+7|\Rightarrow k-7=0\Rightarrow k=7$

Hence, $k=7$