Question

If coordinates of two adjacent vertices of a parallelogram are (3, 2) and (1, 0) and diagonals bisect each other at (-2, 5), find coordinates of the other two vertices?

Answer 1

Hint: First of all draw a parallelogram ABCD and diagonals AC and BD. After that assume the two adjacent coordinates as $\left( {{x}_{C}},{{y}_{C}} \right)\And \left( {{x}_{D}},{{y}_{D}} \right)$. Also, it is given that diagonals bisect each other at the point of bisection is (-2, 5). Now, add the x coordinates of points A and C followed by division with 2 and then equate it to -2 and also add the y coordinates of points A and C followed by division with 2 then equate it to 5. Similarly, do for the coordinates B and D. Now, you will get 4 equations and you have four unknowns ${{x}_{C}},{{y}_{C}},{{x}_{D}},{{y}_{D}}$ so by solving the 4 equations you will get the value of unknowns.

Complete step-by-step solution:
We have given two adjacent vertices of a parallelogram (3, 2) and (1, 0) let us name these vertices as A and B. Also, we have given that the diagonals bisect each other at (-2, 5).
In the below figure, we have drawn a parallelogram ABCD with diagonals AC and BD bisecting each other at point E.

Let us assume the coordinates of point C and D as:
$C\left( {{x}_{C}},{{y}_{C}} \right)\And D\left( {{x}_{D}},{{y}_{D}} \right)$
And marking these coordinates in the above diagram of parallelogram ABCD we get,

Now, we are going to add the x coordinates of vertices A and C and then divide this addition by 2.
$\dfrac{3+{{x}_{C}}}{2}$
As diagonals bisect each other so E is the midpoint of AC and BD. Hence, equating the above to -2 we get,
$\begin{align}
  & \dfrac{3+{{x}_{C}}}{2}=-2 \\
 & \Rightarrow 3+{{x}_{C}}=-4 \\
 & \Rightarrow {{x}_{C}}=-4-3=-7 \\
\end{align}$
Adding the y coordinates of vertices A and C followed by division with 2 we get,
$\dfrac{2+{{y}_{C}}}{2}$
Equating the above to 5 we get,
$\begin{align}
  & \dfrac{2+{{y}_{C}}}{2}=5 \\
 & \Rightarrow 2+{{y}_{C}}=2\left( 5 \right)=10 \\
 & \Rightarrow {{y}_{C}}=10-2=8 \\
\end{align}$
Hence, we have got the coordinates of vertex C as (-7, 8).
Now, we are going to find the coordinates of vertex D as follows:
Adding the x coordinates of vertices B and D and then divide this addition by 2.
$\dfrac{1+{{x}_{D}}}{2}$
Equating the above to -2 we get,
$\dfrac{1+{{x}_{D}}}{2}=-2$
$\begin{align}
  & 1+{{x}_{D}}=2\left( -2 \right)=-4 \\
 & \Rightarrow {{x}_{D}}=-4-1=-5 \\
\end{align}$
Adding the y coordinates of vertices B and D followed by division with 2 we get,
$\dfrac{0+{{y}_{D}}}{2}$
Equating the above to 5 we get,
$\begin{align}
  & \dfrac{{{y}_{D}}}{2}=5 \\
 & \Rightarrow {{y}_{D}}=2\left( 5 \right)=10 \\
\end{align}$
Hence, we got the coordinates of vertex D as (-5, 10).
Hence, we found the coordinates of the remaining vertices (-7, 8) and (-5, 10).

Note: You can check the coordinates of the remaining vertices which we are describing as follows:
Let us check the coordinates of vertex C (-7, 8).
Now, vertex E (-2, 5) is the midpoint of A (3, 2) and C so we are separately finding the midpoint of x and y coordinates and then equate them to -2 and 5 respectively.
$\begin{align}
  & \left( \dfrac{3-7}{2},\dfrac{2+8}{2} \right)=\left( -2,5 \right) \\
 & \Rightarrow \left( \dfrac{-4}{2},\dfrac{10}{2} \right)=\left( -2,5 \right) \\
 & \Rightarrow \left( -2,5 \right)=\left( -2,5 \right) \\
\end{align}$
As you can see that L.H.S is equal to R.H.S so we have found the correct coordinates of vertex C.
Similarly, you can check for vertex D.