Question

If $\left( {2,1} \right),\left( {4,5} \right),\left( { - 1, - 3} \right)$ are the midpoints of the sides of a triangle, then the coordinate of its vertices are
A. $\left( { - 3, - 7} \right),\left( {17,9} \right),\left( {1,1} \right)$
B. $\left( { - 3,7} \right),\left( {7,9} \right),\left( { - 1,1} \right)$
C. $\left( { - 3, - 7} \right),\left( {7,9} \right),\left( {1,1} \right)$
D. None

Answer 1

Hint: In this question, the coordinates of the midpoints of the sides of a triangle are given and the coordinates of the vertices are required. Let the vertices of the triangle be $A\left( {{x_1},{y_1}} \right),B\left( {{x_2},{y_2}} \right)$ and $C\left( {{x_3},{y_3}} \right)$. Then find the coordinates of the midpoints of the sides using the midpoint formula. Thus you’ll get six equations. Solving the equations you’ll obtain the vertices of the triangle.

Formula Used:
Coordinate of midpoint of a line segment joining two points $A\left( {{x_1},{y_1}} \right)$ and $B\left( {{x_2},{y_2}} \right)$ is $\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$

Complete step by step solution:
Let the vertices of the triangle be $A\left( {{x_1},{y_1}} \right),B\left( {{x_2},{y_2}} \right)$ and $C\left( {{x_3},{y_3}} \right)$.
Also, let the midpoints of the sides $AB,BC,CA$ be $D,E,F$ respectively.
The coordinates of the points $D,E,F$ are $\left( {2,1} \right),\left( {4,5} \right),\left( { - 1, - 3} \right)$ respectively.
Using midpoint formula, we get
The coordinate of the midpoint of the side $AB$ be $\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$
So, $\dfrac{{{x_1} + {x_2}}}{2} = 2$ and $\dfrac{{{y_1} + {y_2}}}{2} = 1$
$ \Rightarrow {x_1} + {x_2} = 4.....\left( i \right)$ and ${y_1} + {y_2} = 2.....\left( {ii} \right)$
The coordinate of the midpoint of the side $BC$ be $\left( {\dfrac{{{x_2} + {x_3}}}{2},\dfrac{{{y_2} + {y_3}}}{2}} \right)$
So, $\dfrac{{{x_2} + {x_3}}}{2} = 4$ and $\dfrac{{{y_2} + {y_3}}}{2} = 5$
$ \Rightarrow {x_2} + {x_3} = 8.....\left( {iii} \right)$ and ${y_2} + {y_3} = 10.....\left( {iv} \right)$
The coordinate of the midpoint of the side $CA$ be $\left( {\dfrac{{{x_3} + {x_1}}}{2},\dfrac{{{y_3} + {y_1}}}{2}} \right)$
So, $\dfrac{{{x_3} + {x_1}}}{2} = - 1$ and $\dfrac{{{y_3} + {y_1}}}{2} = - 3$
$ \Rightarrow {x_3} + {x_1} = - 2.....\left( v \right)$ and ${y_3} + {y_1} = - 6.....\left( {vi} \right)$
Solve equations $\left( i \right),\left( {iii} \right),\left( v \right)$ to get the values of ${x_1},{x_2},{x_3}$
Adding the equations $\left( i \right),\left( {iii} \right),\left( v \right)$, we get
$\begin{array}{l}2\left( {{x_1} + {x_2} + {x_3}} \right) = 10\\ \Rightarrow {x_1} + {x_2} + {x_3} = 5.....\left( {vii} \right)\end{array}$
Subtracting equation $\left( i \right)$ from equation $\left( {vii} \right)$, we get ${x_3} = 1$
Subtracting equation $\left( {iii} \right)$ from equation $\left( {vii} \right)$, we get ${x_1} = - 3$
Subtracting equation $\left( v \right)$ from equation $\left( {vii} \right)$, we get ${x_2} = 7$
Now, solve equations $\left( {ii} \right),\left( {iv} \right),\left( {vi} \right)$ to get the values of ${y_1},{y_2},{y_3}$
Adding the equations $\left( {ii} \right),\left( {iv} \right),\left( {vi} \right)$, we get
$\begin{array}{l}2\left( {{y_1} + {y_2} + {y_3}} \right) = 6\\ \Rightarrow {y_1} + {y_2} + {y_3} = 3.....\left( {viii} \right)\end{array}$
Subtracting equation $\left( {ii} \right)$ from equation $\left( {viii} \right)$, we get ${y_3} = 1$
Subtracting equation $\left( {iv} \right)$ from equation $\left( {viii} \right)$, we get ${y_1} = - 7$
Subtracting equation $\left( v \right)$ from equation $\left( {viii} \right)$, we get ${y_2} = 9$
Finally, we have
$\left( {{x_1},{y_1}} \right) = \left( { - 3, - 7} \right),\left( {{x_2},{y_2}} \right) = \left( {7,9} \right),\left( {{x_3},{y_3}} \right) = \left( {1,1} \right)$

Option ‘C’ is correct

Note: There are other methods for solving linear equations. You may use any of the methods. Here we first start by adding all of the equations, then subtracted the original equations from the resulting equation to obtain the values of the unknowns one by one.