Question

The mid-point of the sides of a triangle are at (1,4) , (4,8) and (5,6). Find the coordinates of the vertices of the triangle.

Answer 1

Consider a $\Delta ABC$ with $A(x_1,y_1),B(x_2,y_2)andC(x_3,y_3)$. If $L(1,4),M(4,8)and$$N(5,6)$ are the mid points of $AB,BCandCA$ respectively.

Then,

$1=\frac{x_1+x_2}{2}\Rightarrow x_1+x_2=2......\left(1\right)$

$4=\frac{y_1+y_2}{2}\Rightarrow y_1+y_2=8......\left(2\right)$

$4=\frac{x_2+x_3}{2}\Rightarrow x_2+x_3=8......\left(3\right)$

$8=\frac{y_2+y_3}{2}\Rightarrow y_2+y_3=16.......\left(4\right)$

$5=\frac{x_3+x_1}{2}\Rightarrow x_3+x_1=10......\left(5\right)$

$6=\frac{y_3+y_1}{2}\Rightarrow y_3+y_1=12......\left(6\right)$

On adding (1),(3) and (5) to, we get

$2(x_1+x_2+x_3)=2+8+10$

$x_1+x_2+x_3=10......\left(7\right)$

On adding (2),(4) and (6) to, we get

$2(y_1+y_2+y_3)=8+16+12$

$y_1+y_2+y_3=18......\left(8\right)$

From (1) and (7) to$\Rightarrow$ $x_3=8$

From (3) and (7) to$\Rightarrow$ $x_1=2$

From (5) and (7) to$\Rightarrow$ $x_2=0$

Now,

From (2) and (8) to$\Rightarrow$ $y_3=10$

From (4) and (8) to$\Rightarrow$ $y_1=2$

From (6) and (8) to$\Rightarrow$ $y_2=6$

Hence the point $A(2,2),B(0,6)andC(8,10).$