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In each of the following find the value of $k$ for which the points are collinear.
(i) $(7, -2), (5, 1), (3, k)$
(ii) $(8, 1), (k, -4), (2, -5)$
To do:
We have to find the value of $k$ in each of the given case.
Solution:
(i) Let $A (7, -2), B (5, 1)$ and $C (3, k)$ be the vertices of $\triangle ABC$.
We know that,
If the points $A, B$ and $C$ are collinear then the area of $\triangle ABC$ is zero.
Area of a triangle with vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ is given by,
Area of $\Delta=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$
Therefore,
Area of triangle $ABC=\frac{1}{2}[7(1-k)+5(k+2)+3(-2-1)]$
$0=\frac{1}{2}[7-7k+5k+10+3(-3)]$
$0(2)=(-2k+17-9)$
$0=-2k+8$
$2k=8$
$k=\frac{8}{2}$
$k=4$
The value of $k$ is $4$.
(ii) Let $A (8, 1), B (k, -4)$ and $C (2, -5)$ be the vertices of $\triangle ABC$.
We know that,
If the points $A, B$ and $C$ are collinear then the area of $\triangle ABC$ is zero.
Area of a triangle with vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ is given by,
Area of $\Delta=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$
Therefore,
Area of triangle $ABC=\frac{1}{2}[8(-4+5)+k(-5-1)+2(1+4)]$
$\Rightarrow 0=8-6 k+10$
$\Rightarrow 0=18-6 k$
$\Rightarrow 6k=18$
$\Rightarrow k=\frac{18}{6}$
$\Rightarrow k=3$
The value of $k$ is $3$.
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