Question

Show that four points (0, – 1), (6, 7), (–2, 3) and (8, 3) are the vertices of a rectangle. Also, find its area. [3 MARKS]

Answer 1

Formula: 1 Mark
Application: 1 Mark
Answer: 1 Mark

Let A (0 , -1) , B (6 , 7) , C ( -2 ,3) and D (8 , 3) be the given points. Then,

$AB=\sqrt{{(6-0)}^2+{(7+1)}^2}=\sqrt{36+64}=10$

$CD=\sqrt{{(8+2)}^2+{(3-3)}^2}=\sqrt{100+0}=10$

$AD=\sqrt{{(8-0)}^2+{(3+1)}^2}=\sqrt{64+16}=4\sqrt{5}$

$BC=\sqrt{{(6+2)}^2+{(7-3)}^2}=\sqrt{64+16}=4\sqrt{5}$

$\therefore$ AB = CD and AD = BC

So, ABCD is a parallelogram

Now , $AC=\sqrt{{(-2-0)}^2+{(3-(-1)}^2}=\sqrt{4+16}=2\sqrt{5}$

and , $BD=\sqrt{{(8-6)}^2+{(3-7)}^2}=\sqrt{4+16}=2\sqrt{5}$

Clearly, opposite sides are equal and diagonals are equal
.
Hence, ABCD is a rectangle.

Now, Area of rectangle $ABCD=AB\times BC$

$AB\times BC=10\times 4\sqrt{5}sq.units$

$\Rightarrow AB\times BC=40\sqrt{5}sq.units$