Show that the points (3, 3), (6, 0), (3,-3) and (0, 0) are the vertices of square,
Answers
Answer:
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Step-by-step explanation:
Given :-
The points (3, 3), (6, 0), (3,-3) and (0, 0)
To find :-
Show that the points are the vertices of square?
Solution :-
Given points are (3, 3), (6, 0), (3,-3) and (0, 0)
To prove that the points are the vertices of a square then we have to prove that the lengths of the four points are equal and The lengths of the diagonals are equal.
Let A = (3,3)
Let B = (6,0)
Let C = (3,-3)
Let D = (0,0)
Finding the length of AB :-
Let (x1, y1) = (3,3) => x1 = 3 and y1 = 3
Let (x2, y2) = (6,0) => x2 = 6 and y2 = 0
We know that
The distance between two points (x1, y1) and (x2, y2) is √[(x2-x1)²+(y2-y1)²] units
The distance between two points A and B
=> AB = √[(6-3)²+(0-3)²]
=> AB = √[3²+(-3)²]
=> AB = √(9+9)
=> AB = √18 units ---------------(1)
Finding the length of BC :-
Let (x1, y1) = (6,0) => x1 = 6 and y1 = 0
Let (x2, y2) = (3,-3) => x2 = 3 and y2 = -3
We know that
The distance between two points (x1, y1) and (x2, y2) is √[(x2-x1)²+(y2-y1)²] units
The distance between two points B and C
=> BC = √[(3-6)²+(-3-0)²]
=> BC = √[(-3)²+(-3)²]
=> BC = √(9+9)
=> BC = √18 units ---------------(2)
Finding the length of CD :-
Let (x1, y1) = (3,-3) => x1 = 3 and y1 = -3
Let (x2, y2) = (0,0) => x2 = 0 and y2 = 0
We know that
The distance between two points (x1, y1) and (x2, y2) is √[(x2-x1)²+(y2-y1)²] units
The distance between two points C and D
=> CD = √[(0-3)²+(0-(-3))²]
=> CD = √[(-3)²+(3)²]
=> CD = √(9+9)
=> CD = √18 units ---------------(3)
Finding the length of AD :-
Let (x1, y1) = (3,3) => x1 = 3 and y1 = 3
Let (x2, y2) = (0,0) => x2 = 0 and y2 = 0
We know that
The distance between two points (x1, y1) and (x2, y2) is √[(x2-x1)²+(y2-y1)²] units
The distance between two points A and D
=> AD = √[(0-3)²+(0-3)²]
=> AD = √[3²+(-3)²]
=> AD = √(9+9)
=> AD = √18 units ---------------(4)
Finding the length of AC :-
Let (x1, y1) = (3,3) => x1 = 3 and y1 = 3
Let (x2, y2) = (3,-3) => x2 = 3 and y2 = -3
We know that
The distance between two points (x1, y1) and (x2, y2) is √[(x2-x1)²+(y2-y1)²] units
The distance between two points A and C
=> AC = √[(3-3)²+(-3-3)²]
=> AC = √[0²+(-6)²]
=> AC = √(0+36)
=> AC = √36
=> AC = 6 units ---------------(5)
Finding the length of BD:-
Let (x1, y1) = (6,0) => x1 = 6 and y1 = 0
Let (x2, y2) = (0,0) => x2 = 0 and y2 = 0
We know that
The distance between two points (x1, y1) and (x2, y2) is √[(x2-x1)²+(y2-y1)²] units
The distance between two points B and D
=> BD = √[(0-6)²+(0-0)²]
=> BD = √[(-6)²+(0)²]
=> BD = √(36+0)
=> BD = √36
=> BD = 6 units ---------------(6)
From (1),(2),(3),&(4)
AB = BC = CD = AD
All sides are equal.
From (5)&(6)
AC = BD
diagonals are equal.
Therefore, Given points are the vertices of the square ABCD.
Hence, Proved.
Answer:-
Given points (3, 3), (6, 0), (3,-3) and (0, 0) are the vertices of a square .
Used Formulae :-
→ The distance between two points (x1, y1) and (x2, y2) is √[(x2-x1)²+(y2-y1)²] units
Used Concept :-
→ All sides are equal in a square.
→ Diagonals are equal in a square.