Find the value of $k$ if points $(k, 3), (6, -2)$ and $(-3, 4)$ are collinear.

Given:

Points $(k, 3), (6, -2)$ and $(-3, 4)$ are collinear.

To do:

We have to find the value of $k$.

Solution:

Let $A(k, 3), B(6, -2)$ and $C(-3, 4)$ be the vertices of $\triangle ABC$.

We know that,

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 \( ABC=\frac{1}{2}[k(-2-4)+6(4-3)+(-3)(3+2)] \)

\( 0=\frac{1}{2}[k(-6)+6(1)+(-3)(5)] \)

\( 0(2)=(-6k+6-15) \)

\( 0=-6k-9 \)

\( 6k=-9 \)

\( k=\frac{-9}{6} \)

\( k=\frac{-3}{2} \)

The value of $k$ is $\frac{-3}{2}$.  

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Updated on: 10-Oct-2022

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