Question

Find the value of K if the point A $(2,3)$ ,B$(4,K)$and C$(6,-3)$ are collinear?

Answer 1

Given that the points $A(2,3),B(4,k)$ and $C(6,-3)$ are collinear.

As we know that if three points are collinear then they will lie of a same plane and thus will not form a triangle.

Therefore,

Area of triangle formed by these points will be $0$

Therefore,

Now,

Area of $△$ formed by these points $=\frac{1}{2}\times \left|\begin{array}{ccc}2& 3& 1\\ 4& k& 1\\ 6& -3& 1\end{array}\right|$

Therefore,

$\left|\begin{array}{ccc}2& 3& 1\\ 4& k& 1\\ 6& -3& 1\end{array}\right|=0$

$[2(k-(-3))-3(4-6)+1((-12)-6k)]=0$

$2k+6+6-12-6k=0$

$-4k=0$

$\Rightarrow k=0$

Thus the value of $k$ is $0$.

Hence the correct answer is $0$.