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.
Given:
The area of triangle $ABC$ formed by the points $A (a, 2a), B (-2, 6)$ and $C (3, 1)$ is 10 square units.
To do:
We have to find the value of $a$.
Solution:
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 \( ABC=\frac{1}{2}[a(6-1)-2(1-2a)+3(2a-6)] \)
\( 10=\frac{1}{2}[5a-2+4a+6a-18] \)
\( 10(2)=(15a-20) \)
\( 20=5(3a-4) \)
\( 3a-4=4 \)
\( 3a=4+4 \)
\( 3a=8 \)
\( a=\frac{8}{3} \)
The value of $a$ is $\frac{8}{3}$.
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