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

A B C D

is a rectangle formed by joining the points

A (- 1, - 1)

,

B (- 1, 4)

,

C (5, 4)

, and

D (5, - 1)

.

P, Q, R and S

are the mid-points of sides

A B, B C, C D

and

D A

respectively. Is the quadrilateral

P Q R S

a square? a rectangle? or a rhombus? Justify your answer.

Accepted Answer

We have a rectangle $A B C D$ formed by joining the points $A (- 1, - 1), B (- 1, 4), C (5, 4), and D (5, - 1)$ .

The mid-points of the sides $A B, B C, C D, and D A are P, Q, R, S$ respectively.

We have to find that whether $P Q R S$ is a square, rectangle or rhombus.

In general to find the mid-point $P (x, y)$ of two points $A (x_{1}, y_{1})$ and $A B C D$ we use the formula as given below,

$P (x, y) = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$ .

Co-ordinates of $P$ are $(\frac{- 1 - 1}{2}, \frac{- 1 + 4}{2}) = (- 1, \frac{3}{2})$ .

Similarly, mid-point $Q$ of side $B C$ can be written as, $(\frac{5 - 1}{2}, \frac{4 + 4}{2}) = (2, 4)$

So co-ordinates of $Q$ are $(2, 4)$

Similarly, mid-point $R$ of side $C D$ can be written as, $(\frac{5 + 5}{2}, \frac{4 - 1}{2}) = (5, \frac{3}{2})$

So co-ordinates of $R$ are $(5, \frac{3}{2})$

Similarly, mid-point $S$ of side $D A$ can be written as, $(\frac{5 - 1}{2}, \frac{- 1 - 1}{2}) = (2, - 1)$

So co-ordinates of $S$ are $(2, - 1)$

Finding the side lengths of quadrilateral $P Q R S$ .

$P Q = \sqrt{{(2 + 1)}^{2} + {(4 - \frac{3}{2})}^{2}} = \sqrt{9 + \frac{25}{4}} = \frac{\sqrt{61}}{2}$ units

$Q R = \sqrt{{(2 - 5)}^{2} + {(4 - \frac{3}{2})}^{2}} = \sqrt{9 + \frac{25}{4}} = \frac{\sqrt{61}}{2}$ units

$R S = \sqrt{{(5 - 2)}^{2} + {(\frac{3}{2} + 1)}^{2}} = \sqrt{9 + \frac{25}{4}} = \frac{\sqrt{61}}{2}$ units

$S P = \sqrt{{(- 1 - 2)}^{2} + {(\frac{3}{2} + 1)}^{2}} = \sqrt{9 + \frac{25}{4}} = \frac{\sqrt{61}}{2}$ units

All the sides of quadrilateral are equal.

So now we will check the lengths of the diagonals,

$P R = \sqrt{{(5 + 1)}^{2} + {(\frac{3}{2} - \frac{3}{2})}^{2}}$ $= 6$ units

$Q S = \sqrt{{(2 - 2)}^{2} + {(4 + 1)}^{2}}$ $= 5$ units

All the sides are equal but the diagonals are unequal.

Hence $P Q R S$ is a rhombus.

$A B C D$ is a rectangle formed by joining the points $A (- 1, - 1)$ , $B (- 1, 4)$ , $C (5, 4)$ , and $D (5, - 1)$ . $P, Q, R and S$ are the mid-points of sides $A B, B C, C D$ and $D A$ respectively. Is the quadrilateral $P Q R S$ a square? a rectangle? or a rhombus? Justify your answer.

Important Questions on Co-Ordinate Geometry

Learn and Prepare for any exam you want !

ABCD is a rectangle formed by joining the points A(−1,−1), B(−1,4), C(5,4), and D(5,−1). P, Q, R and S are the mid-points of sides AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.

Important Questions on Co-Ordinate Geometry

Learn and Prepare for any exam you want !

$A B C D$ is a rectangle formed by joining the points $A (- 1, - 1)$ , $B (- 1, 4)$ , $C (5, 4)$ , and $D (5, - 1)$ . $P, Q, R and S$ are the mid-points of sides $A B, B C, C D$ and $D A$ respectively. Is the quadrilateral $P Q R S$ a square? a rectangle? or a rhombus? Justify your answer.