Prove that the points A (7, 3), B (5, 10), C (15, 8) and D (3, 5) taken in order are the vertices of a parallelogram.
Answers
Step-by-step explanation:
Correction:-
The points A (-7, - 3), B (5, 10), C (15, 8) and D (3, -5)
Given :-
The points A (-7, -3), B (5, 10), C (15, 8) and
D (3, -5)
To find:-
Prove that the points A (-7, -3), B (5, 10), C (15, 8) and D (3,- 5) taken in order are the vertices of a parallelogram.
Solution:-
Method-1:-
Given points are A (-7, -3), B (5, 10), C (15, 8)
and D (3, -5)
To prove that the given points A,B,C,D are the vertices of a Parallelogram then we have to prove that Two pair of parallel sides are equal.
AB = CD and BC = DA
We know that
The distance between two points (x1, y1) and
(x2, y2) is √[(x2-x1)^2+(y2-y1)^2] units
Distance between A and B :-
Let (x1, y1)=A(-7,-3) => x1=-7 and y1 = -3
Let (x2, y2)=B(5,10)=>x2=5 and y2=10
Distance between A and B
=>√[(5+7^2+(10+3)^2]
=> √[(12)^2+13^2]
=> √(144+169)
=> √313
AB =√313 units --------------(1)
Distance between B and C:-
Let (x1, y1)=B(5,10) => x1=5 and y1 = 10
Let (x2, y2)=C(15,8)=>x2=15 and y2=8
Distance between B and C
=>√[(15-5)^2+(8-10)^2]
=> √[(10)^2+(-2)^2]
=> √(100+4)
=> √104
BC =√104 units --------------(2)
Distance between C and D :-
Let (x1, y1)=C(15,8) => x1=15 and y1 = 8
Let (x2, y2)=D(3,-5)=>x2=3 and y2=-5
Distance between C and D
=>√[(3-15)^2+(-5-8)^2]
=> √[(-12)^2+(-13)^2]
=> √(144+169)
=> √313 units
CD =√313 units --------------(3)
Distance between D and A :-
Let (x1, y1)=D(3,-5) => x1=3 and y1 = -5
Let (x2, y2)=A(-7,-3)=>x2=-7 and y2=-3
Distance between D and A
=>√[(-7-3)^2+(-3+5)^2]
=> √[(-10)^2+2^2]
=> √(100+4)
=> √104 units
DA =√104 units --------------(4)
From (1)&(3)
AB = CD
From (2)&(4)
BC = DA
Two pair of opposite sides are equal.
A,B,C ,D are the vertices of the paralellogram.
Method-2:-
To prove that the given points A,B,C,D are the vertices of a Parallelogram then we have to prove that Mid point of AC = Mid point of BD Since the diagonals bisect each other in a Parallelogram.
We know that
The mid point of a line segment joining the points (x1, y1) and (x2, y2) is ({x1+x2}/2,{y1+y2}/2)
Mid Point of AC:-
Let (x1, y1)=A(-7,-3) => x1=-7 and y1 = -3
Let (x2, y2)=C(15,8)=>x2=15 and y2=8
Mid Point of AC
= ( {-7+15}/2 , {-3+8}/2 )
= (8/2 , 5/2)
= (4,5/2)
Mid Point of AC = (4,5/2)----------(1)
Mid Point of BD :-
Let (x1, y1)=B(5,10) => x1=5 and y1 = 10
Let (x2, y2)=D(3,-5)=>x2=3 and y2=-5
Mid Point of BD
= ( {5+3}/2 , {10-5}/2 )
= (8/2 , 5/2)
= (4,5/2)
Mid Point of BD = (4,5/2)-----------(2)
From (1)&(2)
Mid Point of AC = Mid Point of BD
A,B,C ,D are the vertices of the paralellogram.
Answer:-
The points A (-7, -3), B (5, 10), C (15, 8) and D (3,- 5) taken in order are the vertices of a parallelogram.
Used formulae:-
In a Parallelogram:
- Two pair of opposite sides are equal.
- Diagonals bisect each other.
Distance formula:
The distance between two points (x1, y1) and
(x2, y2) is√[(x2-x1)^2+(y2-y1)^2] units
Mid Point formula :
The mid point of a line segment joining the points (x1, y1) and (x2, y2) is ({x1+x2}/2,{y1+y2}/2)