Question

The three vertices of a parallelogram are (3, 4), (3, 8) and (9, 8). Find the fourth vertex.

Answer 1

Hint: Assume the coordinate of the fourth vertex of parallelogram ABCD to be (x,y). We know that the diagonal of a parallelogram bisects each other. Since ABCD is a parallelogram, the diagonals must bisect each other. We know that the midpoint(x,y) of A $$(x_1,y_1)$$ and B $$(x_2,y_2)$$ is $$x={\displaystyle \frac{x_1+x_2}{2}}$$ , $$y={\displaystyle \frac{y_1+y_2}{2}}$$ . Using the midpoint formula, find the coordinate of the midpoint of the diagonal BD. Similarly, find the midpoint of the coordinate of the diagonal AC. Since the diagonals meet at a point O. So, the midpoint of the diagonal AC and the diagonal BD must coincide. Now, solve it further and get the values of x and y.

Complete step-by-step answer:

Let the coordinate of the fourth vertex D be (x,y).
We know that the diagonals of a parallelogram bisect each other. Since ABCD is a parallelogram, the diagonals must bisect each other.
For diagonal AC, O is its midpoint.
We know the formula that the midpoint(x,y) of A $$(x_1,y_1)$$ and B $$(x_2,y_2)$$ is $$x={\displaystyle \frac{x_1+x_2}{2}}$$ , $$y={\displaystyle \frac{y_1+y_2}{2}}$$ .
As O is the midpoint of the diagonal AC, we can find its coordinates using the midpoint formula.
We have, A = (3,4) and C = (9,8),
O = $$({\displaystyle \frac{3+9}{2}},{\displaystyle \frac{4+8}{2}})$$ = $$(6,6)$$ ……………………….(1)
As O is the midpoint of the diagonal BD, we can find its coordinates using the midpoint formula.
We have, B = (3,8) and D = (x,y),
O = $$({\displaystyle \frac{3+x}{2}},{\displaystyle \frac{8+y}{2}})$$ …………………….(2)
Comparing equation (1) and equation (2), we get
$$6={\displaystyle \frac{3+x}{2}}$$ ………………….(3)
$$6={\displaystyle \frac{8+y}{2}}$$ ………………….(4)
Solving equation (3), we get
$$\begin{array}{rl}& 6={\displaystyle \frac{3+x}{2}}\\ & \Rightarrow 12=3+x\\ & \Rightarrow 9=x\end{array}$$
Solving equation (4), we get
$$\begin{array}{rl}& 6={\displaystyle \frac{8+y}{2}}\\ & \Rightarrow 12=8+y\\ & \Rightarrow 4=y\end{array}$$
The values of x and y are 4 and 9 respectively.
So, D = (9,4).
Hence, the fourth vertex is (9,4).

Note: We can also solve this question using the distance formula.

Let the fourth vertex of the parallelogram be (x,y).
We know that the opposite sides of a parallelogram are equal to each other.
So, AB = CD and BC = AD.
AB = $$\sqrt{{(3-3)}^2+{(4-8)}^2}=4$$ …………………(1)
BC = $$\sqrt{{(3-9)}^2+{(8-8)}^2}=6$$ ……………………..(2)
CD = $$\sqrt{{(9-x)}^2+{(8-y)}^2}=\sqrt{x^2+y^2-18x-16y+145}$$ ………………………(3)
AD = $$\sqrt{{(3-x)}^2+{(4-y)}^2}=\sqrt{x^2+y^2-6x-8y+25}$$ ……………………..(4)
From equation (1) and equation (3), we get
$$A{B}^2=C{D}^2$$
$$4^2=x^2+y^2-18x-16y+145$$ …………………..(5)
From equation (2) and equation (4), we get
$$B{C}^2=A{D}^2$$
$$6^2=x^2+y^2-6x-8y+25$$ …………………..(6)
We have two equations and two variables. On solving equation (5) and equation (6), we get
x=9 and y=4.
Hence, the fourth vertex is (9,4).