Question

Find the vertices of the triangle whose mid point of sides are $(3,1),(5,6)$ and $(-3,2)$

• A. $(-1,7)(-5,-3)(6,5)$
• B. $(7,1)(2,3)(4,1)$
• C. $(-1,7)(-5,-3)(11,5)$
• D. $(1,7)(5,3)(-11,5)$

Answer 1

The correct option is C $(-1,7)(-5,-3)(11,5)$

Let coordinates of the vertices of the triangle $A,B$ and $C$ be

$A(x_1,y_1)B(x_2,y_2)$ and $C(x_3,y_3)$, respectively.

Now, $\frac{x_2+x_3}{2}=3\Rightarrow x_2+x_3=6.....\left(I\right)$

$\frac{y_2+y_3}{2}=1\Rightarrow y_2+y_3=2......\left(II\right)$

$x_3+x_1=10.....\left(III\right)$

$y_1+y_3=12......\left(IV\right)$

$x_1+x_2=-6......\left(V\right)$

$y_1+y_2=\mathrm{4......}\left(VI\right)$

Adding equations (I), (III) and (V), we get

$2x_1+2x_2+2x_3=10$

$x_1+x_2+x_3=5$

$x_3=11$ ....$[\because x_1+x_2=-6]$

$x_1=-1$ ....$[\because x_2+x_3=6]$

$\therefore x_2=-5$

Adding equations (II), (IV) and (VI), we get

$2y_1+2y_2+2y_3=18$

$y_1+y_2+y_3=9$

$y_3=5$ ....$[\because y_1+y_2=4]$

$y_1=7$ ....$[\because y_2+y_3=2]$

$\therefore y_2=-3$

Therefore, the coordinates of the vertices are $A(-1,7)B(-5,-3)$ and $C(11,5)$.