Question

Prove that A(3, 5), B(6, 0), C(1, -3) and D(-2, 2) are the vertices of a

square.​

Answer 1

Step-by-step explanation:

midpoint of AC(3+1/2,5-3/2)=(2,1)

midpoint of BD(6-2/2,0+2/2)=(2,1)

diagonal bisect each other ABCD is parallelogram

AB=√(6-3)^2+(0-5)^2=√9+25=√34

BC=√(6-1)^2++(0+3)^2=√25+9=√34

since in parallelogram opposite side are equal therefore in ABCD all sides are equal

AC=√(1-3)^2+(-3-5)^2=√4+64=√68

BD=√(-2-6)^2+(2-0)^2=√68

AC=BD

AB^2+BC^2=AC^2

34+34=68

similarly

AD^2+AB^2=BD^2

34+34=68

all angles are 90°

therefore ABCD is a square

Answer 2

Answer:

ABCD is a square.

Step-by-step explanation:

given vertices, A(3, 5), B(6, 0), C(1, -3) and D(-2, 2)

d [A(3, 5),B(6, 0)] =  $\sqrt{(3-6)^2+(5-0)^2}$ = $\sqrt{-3^2 + 5^2}$ = $\sqrt{9+25}$ = $\sqrt{34}$ units

d [B(6, 0),C(1, -3)] =  $\sqrt{(6-1)^2+(0-(-3))^2}$ = $\sqrt{5^2+3^2 }$ = $\sqrt{25+9}$ = $\sqrt{34}$ units

d [C(1, -3),D(-2, 2)] =  $\sqrt{(1-(-2))^2+(-3-2)^2}$ = $\sqrt{3^2 + (-5)^2}$ = $\sqrt{9+25}$ = $\sqrt{34}$ units

d [A(3, 5),D(-2, 2)] =  $\sqrt{(3-(-2))^2+(5-2)^2}$ = $\sqrt{5^2+3^2}$ = $\sqrt{9+25}$ = $\sqrt{34}$ units

from the above calculations, we arrive at a conclusion that

AB = BC = CD = AD

so, all 4 sides are equal  --------------- (1)

also,

d [A(3, 5),C(1, -3)] =  $\sqrt{(3-1)^2+(5-(-3))^2}$ = $\sqrt{2^2+8^2 }$ = $\sqrt{4+64}$ = $\sqrt{68}$ units

d [B(6, 0),D(-2, 2)] =  $\sqrt{(6-(-2))^2+(0-2)^2}$ = $\sqrt{8^2 + (-2)^2}$ = $\sqrt{4+64}$ = $\sqrt{68}$ units

and AC = BD

so, the diagonals are also equal ---------(2)

by (1) &(2), we can say that ABCD is a square.