Question

Find the value of k so that point colinear(7,-2)(5,1)(3,k).

Answer 1

Points are collinear

$(7,-2)(5,1)(3,k)$

Area of $\Delta =0$

$=\frac{1}{2}\left|\begin{array}{ccc}7& -2& 1\\ 5& 1& 1\\ 3& k& 1\end{array}\right|=0$

$=\frac{1}{2}\left[7\right(1-k)+2(5-3)+1(5k-3\left)\right]=0$

$=7(1-k)+4+5k-3=0$

$=7-7k+1+5k=0$

$=8-2k=0$

$k=4$.