Show that, if the points (-2, -7), (2,- 4), (-1,0) and (-5, -3) are joined in order a square is produced. Find the length of each diagonal of the square.
Answers
Step-by-step explanation:
Given :-
The points (-2, -7), (2,- 4), (-1,0) and (-5, -3)
To find :-
i) Show that, the points are the vertices of a square ?
ii) Find the length of each diagonal of the square.
Solution :-
Given points are : (-2, -7), (2,- 4), (-1,0) and (-5, -3)
Let A = (-2, -7)
Let B = (2,- 4)
Let C = (-1,0)
Let D = (-5, -3)
To show that the points A,B,C,D are the vertices of a square then we have to prove that
AB = BC = CD = DA and AC = BD.
Length of AB :-
Let (x1, y1)=A(-2,-7)=>x1= -2 and y1 = -7
Let (x2, y2)=B(2,-4)=> x2 = 2 and y2 = -4
We know that
The distance between two points (x1, y1) and
(x2, y2) is √[(x2-x1)²+(y2-y1)²] units
=>AB = √[(2-(-2))²+(-4-(-7))²]
=> AB =√[(2+2)²+(-4+7)²]
=> AB =√(4²+3²)
=> AB = √(16+9)
=>AB = √25
=> AB = 5 units
Length of AB = 5 units -------(1)
Length of BC :-
Let (x1, y1)=B(2,-4)=>x1= 2 and y1 = -4
Let (x2, y2)=C(-1,0)=> x2 = -1and y2 = 0
We know that
The distance between two points (x1, y1) and
(x2, y2) is √[(x2-x1)²+(y2-y1)²] units
=> BC = √[(-1-2)²+(0-(-4))²]
=> BC =√[(-3)²+(0+4)²]
=>BC =√(9+16)
=>BC= √25
=> BC = 5 units
Length of BC= 5 units -----------(2)
Length of CD :-
Let (x1, y1)=C(-1,0)=>x1= -1 and y1 = 0
Let (x2, y2)=D(-5,-3)=> x2 = -5 and y2 = -3
We know that
The distance between two points (x1, y1) and
(x2, y2) is √[(x2-x1)²+(y2-y1)²] units
=>CD = √[(-5-(-1))²+(-3-0)²]
=>CD =√[(-5+1)²+(-3)²]
=>CD =√[(-4)²+(-3²)]
=> CD = √(16+9)
=>CD = √25
=> CD = 5 units
Length of CD = 5 units -------------(3)
Length of DA :-
Let (x1, y1)=D(-5,-3)=>x1= -5 and y1 = -3
Let (x2, y2)=A(-2,-7)=> x2 = -2 and y2 = -7
We know that
The distance between two points (x1, y1) and
(x2, y2) is √[(x2-x1)²+(y2-y1)²] units
=>DA= √[(2-(-5))²+(-7-(-3))²]
=> DA=√[(-2+5)²+(-7+3)²]
=> DA =√[(3²+(-4²)]
=> DA = √(9+16)
=>DA = √25
=> DA= 5 units
Length of DA = 5 units -------------(4)
Length of AC :-
Let (x1, y1)=A(-2,-7)=>x1= -2 and y1 = -7
Let (x2, y2)=C(-1,0)=> x2 = -1 and y2 = 0
We know that
The distance between two points (x1, y1) and
(x2, y2) is √[(x2-x1)²+(y2-y1)²] units
=>AC= √[(-1-(-2))²+(0-(-7))²]
=> AC =√[(-1+2)²+(0+7)²]
=> AC =√(1²+7²)
=> AC = √(1+49)
=>AC = √50
=> AC = √(2×25)
=> AC= 5√2 units
Length of AC = 5√2 units -------------(5)
Length of BD :-
Let (x1, y1)=B(2,-4) =>x1= 2 and y1 = -4
Let (x2, y2)=D(-5,-3)=> x2= -5 and y2 = -3
We know that
The distance between two points (x1, y1) and
(x2, y2) is √[(x2-x1)²+(y2-y1)²] units
=> BD = √[(-5-2)²+(-3-(-4))²]
=> BD =√[(-7)²+(-3+4)²]
=> BD =√(-7)²+(1)²)
=>BD = √(49+1)
=>BD = √50
=> BD= √(5×5×2)units
Length of BD = 5√2 units -----------(6)
From (1),(2),(3),(4)
AB = BC = CD = DA
All sides are equal.
From (5)&(6)
AC = BD
The diagonals are equal.
Answer:-
I) The given points are the vertices of a square
ii) The length of each diagonal is 5√2 units
Used formulae :-
- In a square all sides are equal.
- In a square two diagonals are equal .
- If a side of a square is a units then it's diagonal is √2 a units.
