Find the value of a for which the area of the triangle

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,

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

The three given points are A(a,2a), B(−2,6) and C(3,1). It is also said that the area enclosed by them is 10 square units. Substituting these values in the above mentioned formula we have,

$\Delta=\frac{1}{2}|(a \times 6+(-2) \times 1+3 \times 2 a)-((-2) \times 2 a+3 \times 6+a \times 1)|$

$10=\frac{1}{2}|(6 a-2+6 a)-(-4 a+18+a)|$

$10=\frac{1}{2}|15 a-20|$

$20=|15 a-20|$

$4=|3 a-4|$

We have $|3 a-4|=4$. Hence either

$3 a-4=4$

$3 a=8$

$a=\frac{8}{3}$

Or

$-(3 a-4)=4$

$4-3 a=4$

$a=0$

Hence the values of ' $a$ ' which satisfies the given conditions are $a=0$ $a=\frac{8}{3}$ .