Question

If mid-points of the sides of $\Delta ABC$ are $(1,2),(0,1)$ and $(0,1),$ then find the coordinates of the three vertices of $\Delta ABC.$

Answer 1

Hint: Here we will find the midpoint of given points using related formulas. Here we will find the midpoint for all points that means $AB$ and $BC$ and $CA$ then finally we will find the coordinate points using some rules and some formula. Then finally we will get the vertices of the given points.

Formula used:
Finding the midpoint of given point $ = \dfrac{{\;{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}$

Complete step-by-step answer:
Let $A({x_1},{y_1}),$ $B({x_2},{y_2}),$ and $C({x_3},{y_3}),$ be the vertices of $\Delta ABC,$ let $P(1,2),Q(0, - 1)$ and $R(2, - 1)$ be the mid-points of sides $BC,AC$ and $AB$
Since, $P$ is the mid-point of $AB.$
$A({x_1},{y_1}),$ $B({x_2},{y_2}),$ and $C({x_3},{y_3}),$
$\therefore \dfrac{{{x_1} + {x_2}}}{2} = 1$ and $\dfrac{{{y_1} + {y_2}}}{2} = 2$
$ \Rightarrow {x_1} + {x_2} = 2 - - - - - - (1)$
$ \Rightarrow {y_2} + {y_3} = 4 - - - - - - (2)$
$Q(0,1)$ is the mid-point of $BC$
$\therefore \dfrac{{{x_2} + {x_3}}}{2} = 0$ and $\dfrac{{{y_2} + {y_3}}}{2} = 1$
$ \Rightarrow {x_2} + {x_3} = 0 - - - - - - (3)$
$ \Rightarrow {y_2} + {y_3} = 2 - - - - - - (4)$
$R(1,0)$ is the mid-point of $AC$
$\therefore \dfrac{{{x_3} + {x_1}}}{2} = 1$ and $\dfrac{{({y_3} + {y_1})}}{2} = 0$
$\Rightarrow {x_3} + {x_1} = 2 - - - - - - - (5)$
$\Rightarrow {y_3} + {y_1} = 0 - - - - - - (6)$
On adding equation $1,3\& 5$
$
  2({x_1} + {x_2} + {x_3}) = 4 \\
  {x_1} + {x_2} + {x_3} = \dfrac{4}{2} \\
  {x_1} + {x_2} + {x_3} = 2 - - - - - - (7) \\
 $
On adding equation $2,4\& 6$
$
  2({y_1} + {y_2} + {y_3}) = 6 \\
  {y_1} + {y_2} + {y_3} = 3 - - - - - - (6) \\
 $
Subtracting equation $(1)$ from equation $(7)$
$
  {x_1} + {x_2} + {x_3} = 2 \\
  {x_1} + {x_2}{\text{ = 2}} \\
  {\text{( - ) ( - ) ( - )}} \\
  {\text{ - - - - - - - - - - - - - - - - - - - - }} \\
\Rightarrow {{\text{x}}_3} = 0 \\
 $
Subtracting equation $(3)$ from equation $(7)$
 $
  {x_1} + {x_2} + {x_3} = 2 \\
  {\text{ + }}{{\text{x}}_2}{\text{ + }}{{\text{x}}_3}{\text{ = 2}} \\
  {\text{( - ) ( - ) ( - )}} \\
  {\text{ - - - - - - - - - - - - - - - - - - - - }} \\
\Rightarrow {{\text{x}}_1} = 2 \\
 $
Subtracting equation $(5)$ from equation $(7)$
 \[
  {x_1} + {x_2} + {x_3} = 2 \\
  {{\text{x}}_1}{\text{ + + }}{{\text{x}}_3}{\text{ = 2}} \\
  {\text{( - ) ( - ) ( - )}} \\
  {\text{ - - - - - - - - - - - - - - - - - - - - }} \\
\Rightarrow {x_2} = 0 \\
 \]
Subtracting equation $(2)$ from equation $(8)$
 $
  {y_1} + {y_2} + {y_3} = 3 \\
  {{\text{y}}_1}{\text{ + }}{{\text{y}}_2}{\text{ = 4}} \\
  {\text{( - ) ( - ) ( - )}} \\
  {\text{ - - - - - - - - - - - - - - - - - - - - }} \\
\Rightarrow {y_3} = - 1 \\
 $
Subtracting equation $(4)$ from equation $(8)$
 $
  {y_1} + {y_2} + {y_3} = 3 \\
  {\text{ + }}{{\text{y}}_2}{\text{ + }}{{\text{y}}_3}{\text{ = 2}} \\
  {\text{ ( - ) ( - ) ( - )}} \\
  {\text{ - - - - - - - - - - - - - - - - - - - - }} \\
\Rightarrow {y_1} = 1 \\
 $
Subtracting equation $(6)$ from equation $(8)$
 $
  {y_1} + {y_2} + {y_3} = 3 \\
  {{\text{y}}_1}{\text{ + + }}{{\text{y}}_3}{\text{ = 0}} \\
  {\text{( - ) ( - ) ( - )}} \\
  {\text{ - - - - - - - - - - - - - - - - - - - - }} \\
\Rightarrow {y_2} = 3 \\
 $
The value of ${x_1} = 2,{x_2} = 0$ and ${x_3} = 0$
The value of ${y_1} = 1,{y_2} = 3$ and ${y_3} = - 1$

Hence, the coordinates of the vertices of the $\Delta ABC$ are $A(2,1),B(0,3),C(0, - 1).$

Additional information:
The medial triangle or midpoint triangle of a triangle \[ABC\] is the triangle with vertices at the midpoints of the triangle's sides \[AB,{\text{ }}AC\] and \[BC.\] If you draw lines from each corner (or vertex) of a triangle to the midpoint of the opposite sides, then those three lines meet at a center, or centroid, of the triangle. The centroid is the triangle's center of gravity, where the triangle balances evenly.

Note:
Here Measure the distance between the two end points, and divide the result by \[2.\] This distance from either end is the midpoint of that line. Alternatively, add the two $x$ coordinates of the endpoints and divide by $2.$Do the same for the $y$ coordinates.