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

Show that the points

(- 4, - 1), (- 2, - 4), (4,0) and (2,3)

are the vertices of a rectangle.

Accepted Answer

The four points are $A (- 4, - 1), B (- 2, - 4), C (4,0) and D (2,3) .$

The distance between two points $(x_{1}, y_{1}) and (x_{2}, y_{2})$ is given by $\sqrt{{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}}$

$A B = \sqrt{{(- 4 + 2)}^{2} + {(- 1 + 4)}^{2}}$

$= \sqrt{4 + 9}$

$= \sqrt{13}$ units

$C D = \sqrt{{(4 - 2)}^{2} + {(0 - 3)}^{2}}$

$= \sqrt{4 + 9}$

$= \sqrt{13}$ units

We have one pair of opposite sides equal.

Now, let us check the other pair of opposite sides.

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

$= \sqrt{36 + 16}$

$= \sqrt{52}$ units

$A D = \sqrt{{(- 4 - 2)}^{2} + {(- 1 - 3)}^{2}}$

$= \sqrt{36 + 16}$

$= \sqrt{52}$ units

The other pair of opposite sides are also equal.

So, the quadrilateral formed by these four points is definitely a parallelogram.

$A C = \sqrt{{(- 4 - 4)}^{2} + {(- 1 - 0)}^{2}}$

$= \sqrt{64 + 1}$

$= \sqrt{65}$ units

$B D = \sqrt{{(- 2 - 2)}^{2} + {(- 4 - 3)}^{2}}$

$= \sqrt{16 + 49}$

$= \sqrt{65}$ units

Now since the diagonals are also equal, we can say that the parallelogram is definitely a rectangle.

Hence, we have proved that the quadrilateral formed by the four given points is a rectangle

Show that the points $(- 4, - 1), (- 2, - 4), (4,0) and (2,3)$ are the vertices of a rectangle.

Important Questions on Co-Ordinate Geometry

Learn and Prepare for any exam you want !

Show that the points (-4,-1),(-2,-4), (4,0) and (2,3) are the vertices of a rectangle.

Important Questions on Co-Ordinate Geometry

Learn and Prepare for any exam you want !

Show that the points $(- 4, - 1), (- 2, - 4), (4,0) and (2,3)$ are the vertices of a rectangle.