Question

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

• A. 183 sq. units
• B. $\frac{183}{2}sq.units.$
• C. 366 sq. units
• D. $\frac{183}{4}sq.units.$

Answer 1

The correct option is B $\frac{183}{2}sq.units.$

Let the given vertices be A(4, -3) and B(-9, 7). Medians CD and BE meet at G. Co-ordinates of G are (1, 4).
Let coordinates of C be (x, y)
So, $\frac{x+4-9}{3}=1\text{and}\frac{y-3+7}{3}=4$
$\Rightarrow x-5=3\Rightarrow y+4=12$
$\Rightarrow x=8\text{and}y=8$
$\therefore$ Co-ordinates of the third vertex are (8, 8).
Thus, the three coordinates of the given triangle are A(4, -3), B(-9, 7) and C(8, 8)
Area of the $\Delta ABC$$=\frac{1}{2}\left[|\right(x_1(y_2-y_3)x_2(y_3-y_1)+x_3(y_1-y_2)\left)|\right]$

$=\frac{1}{2}|4(7-8)-9(8+3)+8(-3-7)|$
$=\frac{1}{2}|-4-99-80|=\frac{1}{2}|-183|$
$=\frac{183}{2}sq.units$