Question

ABC is a triangle whose vertices are A (3, 4), B (-2, -1) and C (5, 3). If G is the centroid and BDCG is a parallelogram then find the coordinates of the vertex D.

• A. (2, 7)
• B. (1, 0)
• C. (5, -6)
• D. (-7, 9)

Answer 1

The correct option is B (1, 0)

Given,
The vertices of a triangle are A(3, 4), B(-2, -1) and C(5, 3)

Centroid of triangle $=[\frac{(x_1+x_2+x_3)}{3},\frac{(y_1+y_2+y_3)}{3}]$
$=[\frac{(3-2+5)}{3},\frac{(4-1+3)}{3}]$
$=(\frac{6}{3},\frac{6}{3})$
$=(2,2)$
The point G is (2, 2)

Let the vertices D be (a, b)
Since BDCG is a parallelogram
Mid-point of BC = Mid-point of DG

$\Rightarrow [\frac{(-2+5)}{2},\frac{(-1+3)}{2}\frac{(2+b)}{2}]$
$\Rightarrow [\frac{3}{2},1]=[\frac{(2+a)}{2},\frac{2+b}{2}]$

Now,
$\Rightarrow \frac{(2+a)}{2}=\frac{3}{2}$
$\Rightarrow 4+2a=6$
$\Rightarrow 2a=2$
$\Rightarrow a=1$

Similarly,
$\Rightarrow \frac{(2+b)}{2}=1$
$\Rightarrow 2+b=2$
$\Rightarrow b=0$

Hence, the vertex of D is (1, 0)