Question

`If $(8,1),(k,-4),(2,-5)$ are collinaer, then $k=$

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is C $3$

Let the given point be $A(8,1),B(K,-4),C(2,-5)$

if the above point are collinear,

they will lie on same line

i.e. they will not form triangle,

Area of $\Delta$ABC$=0$

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

Here,

$\begin{array}{c}{x}_1=8,y_1=1\\ x_2=k,y_2=-4\\ x_3=2,y_3=-5\end{array}$

putting value,

$\begin{array}{c}\Rightarrow \frac{1}{2}\left[8\left(-4+5\right)+k\left(-5-1\right)+2\left(1-4\right)\right]=0\\ \Rightarrow 8\left(-4+5\right)+k\left(-5-1\right)+2\left(1+4\right)=0\times 2\\ \Rightarrow -32+40-5k-k+2+8=0\\ \Rightarrow 8+10-6k=0\\ \Rightarrow -6k=-18\\ \Rightarrow 6k=18\\ \Rightarrow k=\frac{18}{6}\\ \Rightarrow k=3\end{array}$

hence,we get

$k=3$

so, option $C$ is correct.