Question

If the coordinates of the mid-points of the sides of a triangle are$(1,2)(0,-1)and(2,-1)$. Find the coordinates of its vertices.

Answer 1

Let $A(x_1,y_1)$, $B(x_2,y_2)$ and $C(x_3,y_3)$ be the vertices of $△$ABC. Let D(1, 2), E(0, -1), and F(2, -1) be the mid-points of sides BC, CA and AB respectively

Since D is the mid-point of BC

$\therefore \frac{x_2+x_3}{2}=1and\frac{y_2+y_3}{2}=2$

$\Rightarrow x_2+x_3=2and{y}_2+y_3=4$

Similarly, E and F are the mid-points of CA and AB respectively.

$\therefore \frac{x_1+x_3}{2}=0and\frac{y_1+y_3}{2}=-1$

$\Rightarrow x_1+x_3=0and{y}_1+y_3=-2$

$and,\frac{x_1+x_2}{2}=2and\frac{y_1+y_2}{2}=-1$

$\Rightarrow x_1+x_2=4and{y}_1+y_2=-2$

From (i), (ii) and (iii), we get

$(x_2+x_3)+(+x_1{x}_3)+(x_1+x_2)=2+0+4and,(y_2+y_3)+(y_1+y_3)+(y_1+y_2)=4-2-2$

$\Rightarrow 2(x_1+x_2+x_3)=6and2(y_1+y_2+y_3)=0$

$\Rightarrow x_1+x_2+x_3=3and{y}_1+y_2+y_3=0$

From (i) and (Iv), we get

$x_1+2=3and{y}_1+4=0$

$\Rightarrow x_1=1and{y}_1=-4$

So, the coordinates of A are (1,-4)

From (ii) and (iv), we get

$x_2+0=3and{y}_2-2=0$

$\Rightarrow x_2=3and{y}_2=2$

So, coordinates of B are (3, 2)From (iii) and (iv), we get

$x_3+4=3and{y}_3-2=0$

$\Rightarrow x_3=-1and{y}_3=2$

So, coordinates of C are (-1, 2)

Hence, the vertices of the triangle ABC are A(1, -4), B(3, 2) and (-1, 2).