if A (5,0) B (-5,5) C (5,5) FIND D if abcd is a rectangle i will mark u steps
Answers
Step-by-step explanation:
Given :-
ABCD is a rectangle whose vertices are A (5,0), B (-5,5) , C (5,5)
To find :-
Find the fourth vertex D ?
Solution :-
Given that
ABCD is a rectangle whose vertices are A (5,0), B (-5,5) , C (5,5)
Let the coordinates of D be (x,y)
We know that
Opposite sides are equal in a rectangle.
So, AB = CD and BC = DA
Length of AB :-
Let (x1, y1) = (5,0) => x1 = 5 and y1 = 0
Let (x2, y2) = (-5,5) => x2 = -5 and y2 = 5
We know that
The distance between two points (x1, y1) and
(x2, y2) is √[(x2-x1)²+(y2-y1)²] units
The distance between A and B
=> AB =√[(-5-5)²+(5-0)²]
=> AB = √[(-10)²+5²]
=> AB = √(100+25)
=> AB = √125
=> AB = √(5×5×5)
=> AB = 5√5 units
Length of BC :-
Let (x1, y1) = (-5,5) => x1 = -5 and y1 = 5
Let (x2, y2) = (5,5) => x2 = 5 and y2 = 5
We know that
The distance between two points (x1, y1) and
(x2, y2) is √[(x2-x1)²+(y2-y1)²] units
The distance between A and B
=> BC =√[(5+5)²+(5-5)²]
=> BC = √[(10)²+0²]
=> BC = √(100+0)
=> BC = √100
=> BC = 10 units
Length of CD :-
Let (x1, y1) = (5,5) => x1 = 5 and y1 = 5
Let (x2, y2) = (x,y) => x2 = x and y2 = y
We know that
The distance between two points (x1, y1) and
(x2, y2) is √[(x2-x1)²+(y2-y1)²] units
The distance between A and B
=> CD =√[(x-5)²+(y-5)²]
=> CD = √[x²-10x+25+y²-10y+25]
=> CD = √(x²-10x-10y+y²+50) units
Length of DA :-
Let (x1, y1) = (x,y) => x1 = x and y1 = y
Let (x2, y2) = (5,0) => x2 = 5 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 A and B
=> DA =√[(5-x)²+(0-y)²]
=> DA =√[(5-x)²+(-y)²]
=> DA = √[25-10x+x²+y²] units
Now,
AB = CD
=> 5√5 = √(x²-10x-10y+y²+50)
On squaring both sides then
=> (5√5)² = (√(x²-10x-10y+y²+50))²
=> 125 = x²-10x-10y+y²+50
=> x²-10x-10y+y² = 125-50
=> x²-10x-10y+y² = 75 -----------(1)
and BC = DA
=> 10= √[25-10x+x²+y²]
On squaring both sides then
=> 10² = (√[25-10x+x²+y²])²
=> 100 = 25-10x+x²+y²
=> x²+y²-10x = 100-25
=> x²+y²-10x = 75 -----------------(2)
From (1)&(2)
x²-10x-10y+y² = x²+y²-10x
=> -10y = 0
=> y = 0/-10
=> y = 0
On substituting the value of y in (2)
=> x²+0²-10x = 75
=> x²-10x = 75
=> x²-10x-75 = 0
=> x²+5x-15x-75 = 0
=> x(x+5)-15(x+5) = 0
=> (x+5)(x-15) = 0
=> x+5 = 0 or x-15 = 0
=> x = -5 or x = 15
The point = (-5,0) or (15,0)
Answer :-
The fourth vertex is D(-5,0) or D(15,0)
Used formulae:-
→ The distance between two points (x1, y1) and
(x2, y2) is √[(x2-x1)²+(y2-y1)²] units
→ Opposite sides are equal in a rectangle.