Question

plz give me answer this question​

Answer 1

Answer :

They form a Rectangle .

To find :

➡ What type of quadrilateral the points A (2,-2) B (7,3) C (11,-1) D (6,-6)

Solution :

We need to find the length of each side and then we can assume what type of quadrilateral it is ...

Length of AB = ${\sqrt{(x_{2}-x_{1})^2 + (y_{2}-y_{1})^2 }}$

= $\sf{\sqrt{(7-2)^2 + (3-(-2))^2 }}$

= $\sf{\sqrt{(5)^2 + (5)^2 }}$

= $\sf{\sqrt{25 + 25 }}$

= $\sf{\sqrt{50 }}$

.

Length of CD = ${\sqrt{(x_{2}-x_{1})^2 + (y_{2}-y_{1})^2 }}$

= $\sf{\sqrt{(6-11)^2 + (-6-(-1))^2 }}$

= $\sf{\sqrt{(-5)^2 + (5)^2 }}$

= $\sf{\sqrt{25 + 25 }}$

= $\sf{\sqrt{50 }}$

.

Length of BC = ${\sqrt{(x_{2}-x_{1})^2 + (y_{2}-y_{1})^2 }}$

= $\sf{\sqrt{(11-7)^2 + (-1-3))^2 }}$

= $\sf{\sqrt{(4)^2 + (-4)^2 }}$

= $\sf{\sqrt{ 16 + 16 }}$

= $\sf{\sqrt{32 }}$

.

Length of DA = ${\sqrt{(x_{2}-x_{1})^2 + (y_{2}-y_{1})^2 }}$

= $\sf{\sqrt{(2-6)^2 + (-2-(-6))^2 }}$

= $\sf{\sqrt{(-4)^2 + (4)^2 }}$

= $\sf{\sqrt{16 + 16 }}$

= $\sf{\sqrt{32 }}$

.

Length of diagonal AC = ${\sqrt{(x_{2}-x_{1})^2 + (y_{2}-y_{1})^2 }}$

= $\sf{\sqrt{(11-2)^2 + (-1-(-2))^2 }}$

= $\sf{\sqrt{(9)^2 + (1)^2 }}$

= $\sf{\sqrt{81 + 1 }}$

= $\sf{\sqrt{82 }}$

.

Length of diagonal BD = ${\sqrt{(x_{2}-x_{1})^2 + (y_{2}-y_{1})^2 }}$

= $\sf{\sqrt{(6-7)^2 + (-6-3))^2 }}$

= $\sf{\sqrt{(-1)^2 + (-9)^2 }}$

= $\sf{\sqrt{1 + 81 }}$

= $\sf{\sqrt{82 }}$

.

$\implies{Length\ of\ side\ AB\ =\ Length\ of\ side\ CD\ = {\sqrt{50}}}$

$\implies{Length\ of\ side\ BC\ =\ Length\ of\ side\ DA\ = {\sqrt{32}}}$

$\implies{Length\ of\ diagonal\ AC\ =\ Length\ of\ diagonal\ BD\ = {\sqrt{82}}}$

.

Since, the opposite sides and the diagonals are equal the vertices given form a Rectangle.

.

.

Formula used :

Distance formula = ${\sqrt{(x_{2}-x_{1})^2 + (y_{2}-y_{1})^2 }}$