Question

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

Answer 1

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

`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}`

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

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

∴ AD = BC and AC = BD.

So, ADBC is a parallelogram,

`Now,\text{ }AB\text{ }=sqrt((6-0)^2+(7+1)^2)=\sqrt{36+64}=10`

Clearly, ` AB^2 = AD^2 + DB^2 and CD^2 = CB^2 + BD^2`

Hence, ADBC is a rectangle.

Now, Area of rectangle ADBC = AD × DB

= (4√5 × 2√5 ) sq. units = 40 sq. units