Show that four points $(0, - 1), (6, 7), (- 2, 3) a n d (8, 3)$ are the vertices of a rectangle. Also, find its area.

Open in App

Solution

Let A(0-1), B(6,7), C(-2,3) and D(8,3) be the given points. Then
, $A D = \sqrt{{(8 - 0)}^{2} + {(3 + 1)}^{2}} = \sqrt{64 + 16} = 4 \sqrt{5}$
$B C = \sqrt{{(6 + 2)}^{2} + {(7 - 3)}^{2}} = \sqrt{64 + 16} = 4 \sqrt{5}$
$A C = \sqrt{{(- 2 - 0)}^{2} + {(3 + 1)}^{2}} = \sqrt{4 + 16} = 2 \sqrt{5}$
and,
$B D = \sqrt{{(8 - 6)}^{2} + {(3 - 7)}^{2}} = \sqrt{4 + 6} = 2 \sqrt{5}$
$∴ A D = B C a n d A C = B D$
So, ADBC is a parallelogram
Now
$A B = \sqrt{{(6 - 0)}^{2} + {(7 + 1)}^{2}} = \sqrt{36 + 64} = 10$
and, $C D = \sqrt{{(8 + 2)}^{2} + {(3 - 3)}^{2}} = 10$
Clearly, ${A B}^{2} = {A D}^{2} + {D B}^{2} a n d {C D}^{2} = {C B}^{2} + {B D}^{2}$
Hence, ADBC is a rectangle.
Now
,Area of rectangle $A D B C = A D \times D B = (4 \sqrt{5} \times 2 \sqrt{5})$ sq. units=40sq. units

Suggest Corrections

Similar questions

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]

(0, - 1), (6, 7), (- 2, 3)

(8, 3)

(a, a), (- a, - a)

(- \sqrt{3 a}, \sqrt{3 a})

Join BYJU'S Learning Program