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

Question:

Find the value of a for which the area of the triangle formed by the points A(a, 2a), B(−2, 6) and C(3, 1) is 10 square units.

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 $a=0$ $a=\frac{8}{3}$ .

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