Question

In each of the following find the value of $k$, for which the points are collinear.
(i) $(7,-2)$, $(5,1)$, $(3,k)$
(ii) $(8,1)$, $(k,-4)$, $(2,-5)$

• A. (i) $k=4$
• B. (i) $k=5$
• C. (ii) $k=3$
• D. (ii) $k=2$

Answer 1

Answer: (A), (C)

(i) $k=4$

(ii) $k=3$

Since the given points are collinear, they do not form a triangle, which means area of the triangle is Zero.

Area of a triangle with vertices $(x_1,y_1)$ ; $(x_2,y_2)$ and $(x_3,y_3)$ is $\left|\frac{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}{2}\right|$

1) Substituting the points $(x_1,y_1)=(7,-2)$ ; $(x_2,y_2)=(5,1)$ and $(x_3,y_3)=(3,k)$
In the area formula, we get

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

$\left|\frac{7-7k+5k+10-9}{2}\right|=0$

$\left|\frac{8-2k}{2}\right|=0$

$\Rightarrow 8-2k=0$

$\Rightarrow k=4$

2) Substituting the points $(x_1,y_1)=(8,1)$ ; $(x_2,y_2)=(k,-4)$ and $(x_3,y_3)=(2,-5)$ in the area formula, we get

$\left|\frac{8(-4+5)+k(-5-1)+2(1+4)}{2}\right|=0$

$\left|\frac{8-6k+10}{2}\right|=0$

$\left|\frac{18-6k}{2}\right|=0$

$\Rightarrow 18-6k=0$

$\Rightarrow k=3$