For what value of $a$ the point $(a, 1), (1, -1)$ and $(11, 4)$ are collinear?

Given:

Points $(a, 1), (1, -1)$ and $(11, 4)$ are collinear.

To do:

We have to find the value of $a$.

Solution:

Let $A(a, 1), B(1, -1)$ and $C(11, 4)$ 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 \( ABC=\frac{1}{2}[a(-1-4)+1(4-1)+11(1+1)] \)

\( 0=\frac{1}{2}[a(-5)+1(3)+11(2)] \)

\( 0(2)=(-5a+3+22) \)

\( 0=-5a+25 \)

\( 5a=25 \)

\( a=\frac{25}{5} \)

\( a=5 \)

The value of $a$ is $5$. 

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Updated on: 10-Oct-2022

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