For what value of a the point (a, 1), (1, −1), and
Question:

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

Solution:

The formula for the area ‘ $A$ ‘ encompassed by three points $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ is given by the formula,

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

If three points are collinear the area encompassed by them is equal to 0.

The three given points are A(a, 1), B(1, −1) and C(11, 4). It is also said that they are collinear and hence the area enclosed by them should be 0.

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

$0=\frac{1}{2}|(a \times-1+1 \times 4+11 \times 1)-(1 \times 1+11 \times-1+a \times 4)|$

$0=\frac{1}{2}|(-a+4+11)-(1-11+4 a)|$

$0=\frac{1}{2}|(-a+15)-(-10+4 a)|$

$0=\frac{1}{2}|-a+15+10-4 a|$

$0=\frac{1}{2}|-5 a+25|$

$0=-5 a+25$

$5 a=25$

$a=5$

Hence the value of ‘ $a$ ‘ for which the given points are collinear is $a=5$.