# In each of the following find the value of 'k',

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).

Solution:

The given three points will be collinear if the $\Delta$ formed by them has equal to zero area.

(i) Let $\mathrm{A}(7,-2), \mathrm{B}(5,1)$ and $\mathrm{C}(3, \mathrm{k})$ be the vertices of a triangle.

$\therefore$ The given points will be collinear, if

$\operatorname{ar}(\Delta \mathrm{ABC})=0$

or $7(1-\mathrm{k})+5(\mathrm{k}+2)+3(-2-1)=0$

$\Rightarrow 7-7 k+5 k+10+(-6)-3=0$

$\Rightarrow 17-9+5 k-7 k=0$

$\Rightarrow 8-2 k=0 \Rightarrow 2 k=8$

$\Rightarrow \mathrm{k}=\frac{\mathbf{8}}{\mathbf{2}}=4$

The required value of $\mathrm{k}=4$.

(ii) $\mathrm{A}(8,1), \mathrm{B}(\mathrm{k},-4), \mathrm{C}(2,-5)$ are the given points.

$x_{1}=8, x_{2}=k, x_{3}=2$

$y_{1}=1, y_{2}=-4, y_{3}=-5$

the condition for the three points to be collinear is

$x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)=0$

$8 \times(-4+5)+k \times(-5-1)+2 \times(1+4)=0$

i.e. $8-6 k+10=0$, i.e., $6 k=18$, i.e., $k=3$