Find the type of the quadrilateral, if point A(-4, -2), B(-3, -7), C(3,-2) and D(2, 3) are joined
serially
Answers
Step-by-step explanation:
Given :-
The points are A(-4, -2), B(-3, -7), C(3,-2) and D(2, 3) are joined serially .
To find :-
Find the type of the quadrilateral ?
Solution :-
Given points are A(-4, -2), B(-3, -7), C(3,-2) and D(2, 3)
i)Finding AB :-
Let (x1, y1) =A(-4,-2)=>x1 = -4 and y1 = -2
Let (x2, y2)=B(-3,-7)=> x2 = -3 and y2 = -7
We know that
The distance between two points (x1, y1) and ( x2, y2) is √[(x2-x1)²+(y2-y1)²] units
=> AB = √[(-3-(-4))²+(-7-(-2))²]
=> AB =√[(-3+4)²+(-7+2)²]
=> AB =√[1²+(-5)²]
=> AB =√(1+25)
=> AB = √26 units ----------(1)
Finding BC :-
Let (x1, y1) =B(-3,-7)=>x1 = -3and y1 = -7
Let (x2, y2)=C(3,-2)=> x2 = 3 and y2 = -2
We know that
The distance between two points (x1, y1) and ( x2, y2) is √[(x2-x1)²+(y2-y1)²] units
=> BC = √[(3-(-3))²+(-2-(-7))²]
=> BC =√[(3+3)²+(-2+7)²]
=> BC =√[6²+(-5)²]
=> BC =√(36+25)
=> BC = √61 units ----------(2)
iii) Finding CD :-
Let (x1, y1) =C(3,-2)=>x1 = 3 and y1 = -2
Let (x2, y2)=D(2,3)=> x2 = 2 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 = √[(2-3)²+(3-(-2))²]
=> CD =√[(-1)²+(3+2)²]
=> CD =√[1+(5)²]
=> CD =√(1+25)
=> CD = √26 units ----------(3)
iv) Finding DA :-
Let (x1, y1) =D(2,3)=>x1 = 2 and y1 = 3
Let (x2, y2)=A(-4,-2)=> x2 =-4 and y2 = -2
We know that
The distance between two points (x1, y1) and ( x2, y2) is √[(x2-x1)²+(y2-y1)²] units
=> DA = √[(-4-2)²+(-2-3)²]
=> DA =√[(-6)²+(-5)²]
=> DA =√(36+25)
=> DA = √61 units ----------(4)
v)Finding AC :-
Let (x1, y1) =A(-4,-2)=>x1 = -4 and y1 = -2
Let (x2, y2)=C(3,-2)=> x2 = 3 and y2 = -2
We know that
The distance between two points (x1, y1) and ( x2, y2) is √[(x2-x1)²+(y2-y1)²] units
=> AC = √[(3-(-4))²+(-2-(-2))²]
=> AC =√[(3+4)²+(-2+2)²]
=> AC =√[7²+(0²]
=> AC =√(49+0)
=> AC = √49
=> AC = 7 units ----------(5)
vi)Finding BD :-
Let (x1, y1) =B(-3,-7)=>x1 = -3and y1 = -7
Let (x2, y2)=D(2,3)=> x2 = 2 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 = √[(2-(-3))²+(3-(-7))²]
=> BD =√[(2+3)²+(3+7)²]
=> BD =√[5²+10²]
=> BD =√(25+100)
=> BD = √125
=> BD = 5√5 units ----------(6)
From (1)&(3)
AB = CD
From (2)&(4)
BC = DA
Two pair of Opposite sides are equal.
From (5)&(6)
AC ≠ BD
Diagonals are not equal
So, ABCD is a Parallelogram.
Answer:-
The given points are the vertices of the Parallelogram.
Used formulae:-
1.The distance between two points (x1, y1) and ( x2, y2) is √[(x2-x1)²+(y2-y1)²] units
2. In a Parallelogram, Two pair of opposite sides are equal and diagonals are not equal.
