Name: Area of a Triangle (Coordinate Geometry)
Uploaded: 2019-07-19
Duration: 6 min 48 s
Description: Dozo, Mozo and Genius are looking at a globe and notice the Bermuda Triangle on it. Mozo is curious and wants to know the area of the triangle! Study how the...

Question

Two opposite vertices of a square are

(- 1, 2)

and

(3, 2)

. Find the coordinates of other two vertices.

Accepted Answer

The distance between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is given by the formula

$d = \sqrt{{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}}$

Let the two points which are said to be the opposite vertices of a diagonal of a square be $A (- 1, 2)$ and $C (3, 2)$ .

In a square all the sides are of equal length. The diagonals are also equal to each other. Also in a square the diagonal is equal to $\sqrt{2}$ times the side of the square.

Let us find the distance between them which is the length of the diagonal of the square.

$A C = \sqrt{(- 1 - 3)^{2} + (2 - 2)^{2}}$

$= \sqrt{(- 4)^{2} + (0)^{2}}$

$= \sqrt{16}$

$A C = 4$ units

Now we know that in a square,

Side of the square $= \frac{Diagonal of the square}{\sqrt{2}}$

Substituting the value of the diagonal we found out earlier in this equation we have,

Side of the square $= \frac{4}{\sqrt{2}}$ units

Side of the square $= 2 \sqrt{2}$ units

Now, a vertex of a square has to be at equal distances from each of its adjacent vertices.

Let $P (x, y)$ represent another vertex of the same square adjacent to both $‘ A ’$ and $‘ C ’$ .

$A P = \sqrt{(- 1 - x)^{2} + (2 - y)^{2}}$

$C P = \sqrt{(3 - x)^{2} + (2 - y)^{2}}$

But these two are nothing but the sides of the square and need to be equal to each other.

So, $A P = C P$

$\sqrt{(- 1 - x)^{2} + (2 - y)^{2}} = \sqrt{(3 - x)^{2} + (2 - y)^{2}}$

Squaring on both sides we have,

$1 + x^{2} + 2 x + 4 + y^{2} - 4 y = 9 + x^{2} - 6 x + 4 + y^{2} - 4 y$

$8 x = 8$

$x = 1$

Substituting this value of $‘ x ’$ and the length of the side in the equation for $‘ A P ’$ we have,

$A P = \sqrt{(- 1 - x)^{2} + (2 - y)^{2}}$

$2 \sqrt{2} = \sqrt{(- 1 - 1)^{2} + (2 - y)^{2}}$

$2 \sqrt{2} = \sqrt{(- 2)^{2} + (2 - y)^{2}}$

Squaring on both sides,

$8 = (- 2)^{2} + (2 - y)^{2}$

$8 = 4 + 4 + y^{2} - 4 y$

$0 = y^{2} - 4 y$

We have a quadratic equation. Solving for the roots of the equation we have,

$y^{2} - 4 y = 0$

$y (y - 4) = 0$

The roots of this equation are $0$ and $4$ .

Therefore, the other two vertices of the square are $(1, 0)$ and $(1, 4)$ .

Two opposite vertices of a square are $(- 1, 2)$ and $(3, 2)$ . Find the coordinates of other two vertices.

Important Questions on Co-Ordinate Geometry

Learn and Prepare for any exam you want !

Two opposite vertices of a square are (-1,2) and (3,2). Find the coordinates of other two vertices.

Important Questions on Co-Ordinate Geometry

Learn and Prepare for any exam you want !

Two opposite vertices of a square are $(- 1, 2)$ and $(3, 2)$ . Find the coordinates of other two vertices.