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.

• A. $(1,4),(-3,2),(1-2)$
• B. $(1,-4),(3,2),(-1,2)$
• C. $(1,-4),(-3,2),(1,-2)$
• D. $(-1,4),(3,2),(-1,2)$

Answer 1

Answer: (B)

$(1,-4),(3,2),(-1,2)$

Let $A\left(x_1{y}_1\right),B(x_2,y_2)andC(x_3,y_3)$ be the vertices of $△$ABC.

Let D(1,2) ,E(0,-1) and F(2,-1) be the mid -points of sides BC,AC and AB

Since D is the mid point of BC.

$A\left(x_1{y}_1\right),B(x_2,y_2)andC(x_3,y_3)\therefore \frac{x_2+x_3}{2}=1and\frac{y_2+y_3}{2}=2⟹x_2+x_3=2y_2+y_3=4\to$ (i)

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

$\therefore \frac{x_1+x_3}{2}=0and\frac{y_1+y_3}{2}=-1⟹x_1+x_3=2and{y}_1+y_3-2\to \left(2\right)\therefore \frac{x_1+x_2}{2}=2and\frac{y_1+y_2}{2}=-1⟹x_1+x_2=4and{y}_1+y_2=-2\to \left(3\right)$

From (1),(2),and(3)

$(x_2+x_3)+(x_1+x_3)+(x_1+x_2)=2+0+4(y_2+y_3)+(y_1+y_3)+(y_1+y_2)=4-2-2⟹x_1+x_2+x_3=3and{y}_1+y_2+y_3=0\to \left(4\right)$

From (1) and (4)

$x_1+2=3and{y}_1+4=0⟹x_1=1and{y}_1=-4\therefore (1,-4)$

From (2) and (4)

$x_2+0=3and{y}_2-2=0⟹(x_2,y_2)=B(3,2)$

From (3) and (4)

$x_3+4=3and{y}_3-2=0\therefore (x_3,y_3)=C(-1,2)$

$\therefore$ Coordinate of $△$ABC are $\Rightarrow$ $(1,-4);B(3,2);C(-1,2)$