Question

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

Answer 1

Solution :

Let,

• A( 0 , -1 )
• B( 6 , 7 )
• C( -2 , 3 )
• D( 8 , 3 )

We have Distance Formula,

$\tt\dagger\: \: \:AB = \sqrt{ {(x_2 - x_1)}^{2} + {(y_2 - y_1)}^{2} }$

Let's Find, Distance.

$\rule{90}1$

For Line Segment AB,

$\tt \longmapsto AB = \sqrt{ {(6 - 0)}^{2} + {(7 + 1)}^{2} }$

$\tt\longmapsto AB = \sqrt{ {6}^{2} + {8}^{2} }$

$\tt \longmapsto AB = \sqrt{100}$

$\tt\longmapsto AB = 10 \: units.$

$\rule{90}1$

For Line Segment DC,

$\tt\longmapsto DC = \sqrt{ {(8 + 2)}^{2} +{ (3 - 3) }^{2} }$

$\tt\longmapsto DC = \sqrt{ {10}^{2} }$

$\tt\longmapsto DC =10 \: units.$

$\rule{90}1$

For Line Segment AD,

$\tt\longmapsto AD = \sqrt{ {(8)}^{2} + {(3 + 1)}^{2} }$

$\tt\longmapsto AD = \sqrt{64 + 16}$

$\tt\longmapsto AD = \sqrt{80}$

$\tt\longmapsto AD =4 \sqrt{5} \: units.$

$\rule{90}1$

For Line segment BC,

$\tt\longmapsto AD = \sqrt{ {( - 2 - 6)}^{2} + {(3 - 7)}^{2} }$

$\tt\longmapsto AD = \sqrt{ {(8)}^{2} + {(4)}^{2} }$

$\tt\longmapsto AD = \sqrt{64 + 16}$

$\tt\longmapsto AD =4 \sqrt{5} \: units$

Now,

• AB = DC.
• AD = BC.

We can Clearly see AB = DC and AD = BC with 90° of Angle, Hence ABCD is a Rectangle.

$\rule{200}3$

$\tt\longmapsto{Area \: of \: Rectangle = L \times B = 10 \times 4\sqrt{5}}.$

$\tt\longmapsto{Area \: of \: Rectangle = 40\sqrt{5}\: Sq\: unit.}$

Therefore, the area of Rectangle be 40√5 sq units.

$\rule{90}1$

Note : Graph provide above.