Question

If the mid points of sides of $\Delta ABC$ are $(2,3),(3,7)$ and $(-2,1)$, then the possible coordinates of its vertices are

• A. $(-1,5)$
• B. $(-3,-3)$
• C. $(7,9)$
• D. $(5,-1)$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $(-1,5)$
B $(-3,-3)$
C $(7,9)$

Let $A(x_1,y_1),B(x_2,y_2),C(x_3,y_3)$
$M_1(2,3)$ is midpoint of side $BC$
$M_2(3,7)$ is midpoint of side $AC$
$M_3(-2,1)$ is midpoint of side $AB$

Now, $\frac{x_1+x_2}{2}=-2$
$\Rightarrow x_1+x_2=-4\cdots \left(1\right)$
Similarly, $x_2+x_3=4\cdots \left(2\right)$
$x_1+x_3=6\cdots \left(3\right)$

Adding equation $\left(1\right),\left(2\right)$ and $\left(3\right)$,
$2(x_1+x_2+x_3)=6\Rightarrow x_1+x_2+x_3=3\cdots \left(4\right)$
Using equation $\left(1\right)$ and $\left(4\right),$
$x_3=7$
Using equation $\left(2\right)$ and $\left(4\right),$
$x_1=-1$
Using equation $\left(3\right)$ and $\left(4\right),$
$x_2=-3$

Similarly, for $y-$coordinates
$y_1+y_2=2\cdots \left(5\right)y_2+y_3=6\cdots \left(6\right)y_3+y_1=14\cdots \left(7\right)$

Adding equation $\left(5\right),\left(6\right)$ and $\left(7\right)$
$y_1+y_2+y_3=11\cdots \left(8\right)$
Using equation $\left(5\right)$ and $\left(8\right),$
$y_3=9$
Using equation $\left(6\right)$ and $\left(8\right),$
$y_1=5$
Using equation $\left(7\right)$ and $\left(8\right),$
$y_2=-3$

Hence, the coordinates of the triangle vertex are
$A=(-1,5),B=(-3,-3),C=(7,9)$


Alternate Solution :


Let $A(x_1,y_1),B(x_2,y_2),C(x_3,y_3)$
$M_1(2,3)$ is midpoint of side $BC$
$M_2(3,7)$ is midpoint of side $AC$
$M_3(-2,1)$ is midpoint of side $AB$

We know that, $B{M}_1{M}_2{M}_3$ is a parallelogram.
And its diagonal bisect each other,
$(\frac{x_2+3}{2},\frac{y_2+7}{2})=(\frac{2-2}{2},\frac{3+1}{2})\Rightarrow x_2=-3,y_2=-3\therefore B=(-3,-3)$

Similarly, $C{M}_1{M}_3{M}_2$ is a parallelogram.
$(\frac{x_3-2}{2},\frac{y_3+1}{2})=(\frac{2+3}{2},\frac{3+7}{2})\Rightarrow x_3=7,y_3=9\therefore C=(7,9)$

Similarly, $A{M}_3{M}_1{M}_2$ is a parallelogram.
$(\frac{x_1+2}{2},\frac{y_1+3}{2})=(\frac{3-2}{2},\frac{7+1}{2})\Rightarrow x_1=-1,y_1=5\therefore A=(-1,5)$