Find the value of $k$, if the points $A (7, -2), B (5, 1)$ and $C (3, 2k)$ are collinear.

Given:

Points $A (7, -2), B (5, 1)$ and $C (3, 2k)$ are collinear.

To do:

We have to find the value of $k$.

Solution:

Let $A (7, -2), B (5, 1)$ and $C (3, 2k)$ 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 \( ABC=\frac{1}{2}[7(1-2k)+5(2k+2)+3(-2-1)] \)

\( 0=\frac{1}{2}[7-14k+10k+10+3(-3)] \)

\( 0(2)=(-4k+17-9) \)

\( 0=-4k+8 \)

\( 4k=8 \)

\( k=\frac{8}{4} \)

\( k=2 \)

The value of $k$ is $2$. 

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Updated on: 10-Oct-2022

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