Question

show that the vertices a (3, 5),B (6, 0),C(1, -3) and D(-2, 2) are the vertices of square ABCD​

Answer 1

Answer:

all sides equal then it is square

Step-by-step explanation:

length AB= sqrt((6-3)^2+(0-5)^2)=sqrt(34)

length BC =sqrt((1-6)^2+((-3)-0)^2)=sqrt(34)

length CD = sqrt(((-2)-1)^2+(2-(-3))^2)= sqrt(34)

length DA = sqrt((3-(-2))^2+(5-2)^2)

sqrt(34)

all sides equal

so it is square

Answer 2

Answer:

all sides equal then it is square

Step-by-step explanation:

length AB= sqrt((6-3)^2+(0-5)^2)=sqrt(34)

length BC =sqrt((1-6)^2+((-3)-0)^2)=sqrt(34)

length CD = sqrt(((-2)-1)^2+(2-(-3))^2)= sqrt(34)

length DA = sqrt((3-(-2))^2+(5-2)^2)

sqrt(34)

all sides equal

therfore ther are square as well as rhombus

to prove it as a square we have to prove that its diagonals are equal

i.e AC=AD

AC = $\sqrt(3-1)^{2} + (5+3)x^{2}$$\sqrt(3-1)^{2} +(5+8)x^{2}$

= $\sqrt68$  

BC = $\sqrt(6+2)^{2}+(0-2)x^{2}$

= $\sqrt68$