Aspire Faculty ID #3640 · Topic: NIMCET 2012 · Just now
NIMCET 2012

If $(4,-3)$ and $(-9,7)$ are two vertices of a triangle and $(1,4)$ is its centroid, find the area of the triangle.

Solution

Let the third vertex be $(x,y)$. Centroid formula: $\displaystyle \left(\frac{4 + (-9) + x}{3}, \frac{-3 + 7 + y}{3}\right) = (1,4)$

From x–coordinate: $\dfrac{4 - 9 + x}{3} = 1$ 
$x - 5 = 3$ 
$x = 8$

From y–coordinate: $\dfrac{-3 + 7 + y}{3} = 4$ 
$y + 4 = 12$ 
$y = 8$ 
So third vertex is: $(8,8)$

Vertices: $A(4,-3), B(-9,7), C(8,8)$ 

Area: $\displaystyle \text{Area} = \frac{1}{2}\left| x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) \right|$ Substitute: 
$\displaystyle = \frac{1}{2}\left|4(7-8) + (-9)(8+3) + 8(-3-7) \right|$ 
$\displaystyle = \frac{1}{2}\left| 4(-1) - 9(11) + 8(-10) \right|$ 
$\displaystyle = \frac{1}{2}\left| -4 - 99 - 80 \right|$ 
$\displaystyle = \frac{1}{2} \times 183$ $\displaystyle = \frac{183}{2}$
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