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 of the triangle.

Answer 1

$\textbf{Given:}$

$\text{Mid points of sides of the triangle are (1,2),(0,-1) and (2,-1)}$

$\textbf{To find:}$

$\text{Vertices of the given triangle}$

$\textbf{Solution:}$

$\text{Let the vertices of the triangle be}$

$\text{ A(x_1,y_1) , B(x_2,y_2) and C(x_3,y_3) }$

$\textbf{Mid point of AB=(1,2)}$

$\implies(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2})=(1,2)$

$\implies\dfrac{x_1+x_2}{2}=1\;\text{and}\;\dfrac{y_1+y_2}{2}=2$

$\implies\;x_1+x_2=2\;.......(1)\;\text{and}\;y_1+y_2=4\;......(2)$

$\textbf{Mid point of BC=(0,-1)}$

$\implies(\dfrac{x_2+x_3}{2},\dfrac{y_2+y_3}{2})=(0,-1)$

$\implies\dfrac{x_2+x_3}{2}=0\;\text{and}\;\dfrac{y_2+y_3}{2}=-1$

$\implies\;x_2+x_3=0\;.......(3)\text{and}\;y_2+y_3=-2\;......(4)$

$\textbf{Mid point of AC=(2,-1)}$

$\implies(\dfrac{x_1+x_3}{2},\dfrac{y_1+y_3}{2})=(2,-1)$

$\implies\dfrac{x_1+x_3}{2}=2\;\text{and}\;\dfrac{y_2+y_3}{2}=-1$

$\implies\;x_1+x_3=4\;.......(5)\;\text{and}\;y_1+y_3=-2\;......(6)$

$\text{Adding (1),(3) and (5)}$

$2(x_1+x_2+x_3)=6$

$\implies\;x_1+x_2+x_3=3$.......(7)

$\text{using (1) in (7)}$

$2+x_3=3\implies\,x_3=1$

$\text{using (3) in (7)}$

$x_1+0=3\implies\,x_1=3$

$\text{using (5) in (7)}$

$x_2+4=3\implies\,x_2=-1$

$\text{Adding (2),(4) and (6)}$

$2(y_1+y_2+y_3)=0$

$\implies\;y_1+y_2+y_3=0$.......(8)

$\text{using (2) in (8)}$

$4+y_3=0\implies\,y_3=-4$

$\text{using (4) in (8)}$

$y_1+(-2)=0\implies\,y_1=2$

$\text{using (6) in (8)}$

$y_2+(-2)=0\implies\,y_2=2$

$\bf(x_1,y_1)=(3,2)$

$\bf(x_2,y_2)=(-1,2)$

$\bf(x_3,y_3)=(1,-4)$

$\therefore\textbf{The vertices are (3,2), (-1,2) and (1,-4)}$

Answer 2

Answer:

Step-by-step explanation: