 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 $\Delta ABC.$ Verified
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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 $= \dfrac{{\;{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}$

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 $A(2,1),B(0,3),C(0, - 1).$

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.
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.