Question

show that A (1,0) B (0,1) C (1,2) D ((2,1) are vertices of paralleogram ABCD and find the ABCD is a rectangle

Answer 1

$\pink{\underline{\bold{\large{Answer}}}}$

Your solution with explanation.

A(1,0)
B(0,1)
C(1,2)
D(2,1)

We know that mid point of line segment joining the points (x,y) and (X,Y) is :

$(\frac{X+x}{2},\frac{Y+y}{2})$

Now coordinates of mid point of AC = $\\(\frac{1+1}{2},\frac{0+2}{2})\\\\\implies (1,1)$

coordinate of mid point of BD = $\\(\frac{0+2}{2},\frac{1+1}{2})\\\\\implies(1,1)$

So, Coordinate of mid point of BD = Diagonals of quadrilateral ABCD bisect each other => ABCD is a parallelogram

Now , we shall see whether is ABCD is a rectangle or not :

$AC = \sqrt{(1-1)^{2} + (2-0)^{2}} \\ AC = \sqrt{(0)^{2} + (2)^{2}}\\AC= \sqrt{2^2}\\ AC = 2 units$

$BD = \sqrt{(2-0)^{2} + (1-1)^{2}} \\ AC = \sqrt{(2)^{2} + (0)^{2}}\\AC= \sqrt{2^2}\\ BD = 2 units$

$\boxed{\red{AC = BD}}$

So, ABCD is a rectangle.

Answer 2

Answer:

hope it's help ful

Step-by-step explanation:

thanks