Show that the points A (3, 5), B (6, 0), C (1, -3) and D (-2, 2) are the vertices of a square ABCD
Answers
Answer:
No, the points the vertices of a square ABCD
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Answer:
The points A ( 3 , 5 ), B ( 6 , 0 ), C ( 1 , -3 ) and D ( -2 , 2 ) are the vertices of a square ABCD.
\bullet \bf \ Distance \ formula = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}∙ Distance formula=
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
\bullet \ \sf A B = \sqrt{(6-3)^2+(0-5)^2}∙ AB=
(6−3)
2
+(0−5)
2
\begin{gathered}= \sf \sqrt{3^2+(-5)^2} \\\\= \sqrt{9+25} \\\\= \sqrt{34} \ units\end{gathered}
=
3
2
+(−5)
2
=
9+25
=
34
units
\bullet \sf \ BC = \sqrt{(1-6)^2+(-3-0)^3}∙ BC=
(1−6)
2
+(−3−0)
3
\begin{gathered}= \sf \sqrt{(-5)^2+(-3)^2} \\\\= \sqrt{25+9} \\\\= \sqrt{34} \ units\end{gathered}
=
(−5)
2
+(−3)
2
=
25+9
=
34
units
\bullet \sf \ CD = \sqrt{(-2-1)^2+(2-(-3))^2}∙ CD=
(−2−1)
2
+(2−(−3))
2
\begin{gathered}= \sf \sqrt{(-3)^2+5^2} \\\\= \sqrt{9+25} \\\\= \sqrt{34} \ units\end{gathered}
=
(−3)
2
+5
2
=
9+25
=
34
units
\bullet \sf \ AD = \sqrt{(-2-3)^2+(2-5)^2}∙ AD=
(−2−3)
2
+(2−5)
2
\begin{gathered}= \sf \sqrt{(-5)^2+(-3)^2} \\\\=\sqrt{25+9 } \\\\= \sqrt{34} \ units\end{gathered}
=
(−5)
2
+(−3)
2
=
25+9
=
34
units
∴ AB = BC = CD = AD
That is,
Sides are of equal length.
\bullet \sf \ Diagonal \ AC = \sqrt{(1-3)^2+(-3-5)^2}∙ Diagonal AC=
(1−3)
2
+(−3−5)
2
\begin{gathered}= \sf \sqrt{(-2)^2+(-8)^2} \\\\= \sqrt{4+64} \\\\= \sqrt{68} \ units\end{gathered}
=
(−2)
2
+(−8)
2
=
4+64
=
68
units
\bullet \sf \ Diagonal \ BD = \sqrt{(-2-6)^2+(2-0)^2}∙ Diagonal BD=
(−2−6)
2
+(2−0)
2
\begin{gathered}= \sf \sqrt{(-8)^2+2^2} \\\\=\sqrt{64+4} \\\\= \sqrt{68} \ units\end{gathered}
=
(−8)
2
+2
2
=
64+4
=
68
units
∴ AC = BD
That is,
Diagonals are equal.
Since sides and diagonals are equal, the given points are the vertices of the square ABCD