Question

$(k,k),(2,3)$ and $(4,-1)$ are collimear. So find the value of $k$.

Answer 1

We know that,

Area of $\Delta ABC=0$

$=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]=0$

Here,

$x_1=k,y_1=k$

$\begin{array}{c}{x}_2=2,y_2=3\\ x_3=4,y_3=-1\end{array}$

Putting values,

$=\frac{1}{2}\left[k\left(3-\left(-1\right)\right)+2\left(\left(-1\right)-k\right)+4\left(k-3\right)\right]=0$

$=\frac{1}{2}\left[3k+k-2-2k+4k-12\right]=0$

$=\frac{1}{2}\left[6k-14\right]=0$

$\begin{array}{c}\frac{6k-14}{2}=0\\ 6k-14=0\\ 6k=14\\ k=\frac{14}{6}=\frac{7}{3}\end{array}$

$k=\frac{7}{3}$

Hence, this is the answer.