Question

The mid -points of the sides of a triangle are$(1,5-1),(0,4,-2)and(2,3,4).$ Find its vertices.

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

Answer 1

The correct option is A

$A(1,2,3),B(3,4,5)andC(-1,6,-7).$

Let $A(x_1,y_1,z_1),B(x_2,y_2,z_2)andC(x_3,y_3,z_3)$ be the vertices of the given triangle and let $D(1,5,-1),E(0,4,-2)andF(2,3,4)$ be the mid points of the sides BC, CA and AB respectively.
Now,
D is mid-point of $BC$
$\therefore \frac{x_2+x_2}{2}=1,\frac{y_2+y_3}{2}=5,\frac{z_2+z_3}{2}=-1\Rightarrow x_2+x_3=2,y_2+y_3=10,z_2+z_3=-2$
E is the mid point of CA
$\therefore \frac{x_1+x_3}{2}=0,\frac{y_1+y_3}{2}=4,\frac{z_1+z_3}{2}=-2\Rightarrow x_1+x_3=0,y_1+y_3=8,z_1+z_3=-4$
F is the mid point of AB
$\therefore \frac{x_1+x_2}{2}=2,\frac{y_1+y_2}{2}=3,\frac{z_1+z_2}{2}=4\Rightarrow x_1+x_2=4,y_1+y_2=6,z+1+z_2=8$
Adding first three equations in (i) ,(ii) and (iii) , we obtain
$2(x_1+x_2+x_3)=2+0+4\Rightarrow x_1+x_2+x_3=3$
Solving first three equations in (i) , (ii) and (iii) with
$x_1+x_2+x_3=3,$ we obtain
$x_1=1,x_2=3,x_3=-1.$
Adding next three equations in (i) (ii) and (iii), we obtain
$2(y_1+y_2+y_3)=10+8+6$
$\Rightarrow y_1+y_2+y_3=12$
Solving next three equations in (i) , (ii) and (iii) with $y_1+y_2+y_3=12,$ we obtain
$y_1=2,y_2=4,y_3=6.$
Adding last three equations in (i) ,(ii) and (iii) we obtain
$2(z_1+z_2+z_3)=-2-5+8\Rightarrow z_1+z_2+z_3=1.$
Solving last three equation in (i) , (ii) and (iii) with $z_1+z_2+z_3=1,$ we obtain
$z_1=3,z_2=5,z_3=-7.$
Thus the vertices of the triangle are $A(1,2,3),B(3,4,5)andC(-1,6,-7).$