If the distance between two points A(a,2) and B(2,-2)is 5 units.find the possible values of a.
Answers
Step-by-step explanation:
Given :-
The distance between two points A(a,2) and B(2,-2) is 5 units
To find :-
The possible values of a .
Solution :-
Given points are A(a,2) and B(2,-2)
Let (x₁ , y₁) = (a,2) => x₁ = a and y₁= = 2
Let (x₂,y₂)= (2,-2) => x₂= 2 and y₂ = -2
We know that
The distance between two points (x₁ , y₁) and (x₂,y₂) = is √[(x₂-x₁)²+(y₂-y₁)²] units
Now,
The distance between two points A and B
=> AB = √[(2-a)²+(-2-2)²] units
=> AB = √[(2-a)²+(-4)²]
=> AB = √[2²-(2×2×a)+a²+16]
=> AB = √(4-4a+a²+16)
=> AB = √(a²-4a+20) units
According to the given problem
The distance between two points A and B = 5 units
=> √(a²-4a+20) = 5
On squaring both sides then
=> [√(a²-4a+20)]² = 5²
=> a²-4a+20 = 25
=> a²-4a+20-25 = 0
=> a²-4a-5 = 0
=> a²+a-5a-5 = 0
=> a(a+1) -5(a+1) = 0
=> (a+1)(a-5) = 0
=> a+1 = 0 (or) a-5 = 0
=> a = -1 (or) a = 5
Therefore, a = -1 and 5
Answer :-
The possible values of a for the given problem are -1 and 5
Check :-
Case -1 :-
If a = -1 then ,the distance between A and B
=> AB = √[(2+1)²+(-2-2)²] units
=> AB = √[3²+(-4)²]
=> AB = √(9+16)
=> AB = √25
=> AB = ± 5 units
Therefore, AB = 5 units
Since, Distance can't be negative.
Case -2:-
If a = 5 then ,the distance between A and B
=> AB = √[(2-5)²+(-2-2)²] units
=> AB = √[(-3)²+(-4)²]
=> AB = √(9+16)
=> AB = √25
=> AB = ± 5 units
Therefore, AB = 5 units
Since, Distance can't be negative.
Verified the given relations in the given problem.
Used formulae:-
The distance between two points (x₁ ,y₁) and (x₂,y₂) is √[(x₂-x₁)²+(y₂-y₁)²] units
Step-by-step explanation:
Step-by-step explanation:
Given :-
The distance between two points A(a,2) and B(2,-2) is 5 units
To find :-
The possible values of a .
Solution :-
Given points are A(a,2) and B(2,-2)
Let (x₁ , y₁) = (a,2) => x₁ = a and y₁= = 2
Let (x₂,y₂)= (2,-2) => x₂= 2 and y₂ = -2
We know that
The distance between two points (x₁ , y₁) and (x₂,y₂) = is √[(x₂-x₁)²+(y₂-y₁)²] units
Now,
The distance between two points A and B
=> AB = √[(2-a)²+(-2-2)²] units
=> AB = √[(2-a)²+(-4)²]
=> AB = √[2²-(2×2×a)+a²+16]
=> AB = √(4-4a+a²+16)
=> AB = √(a²-4a+20) units
According to the given problem
The distance between two points A and B = 5 units
=> √(a²-4a+20) = 5
On squaring both sides then
=> [√(a²-4a+20)]² = 5²
=> a²-4a+20 = 25
=> a²-4a+20-25 = 0
=> a²-4a-5 = 0
=> a²+a-5a-5 = 0
=> a(a+1) -5(a+1) = 0
=> (a+1)(a-5) = 0
=> a+1 = 0 (or) a-5 = 0
=> a = -1 (or) a = 5
Therefore, a = -1 and 5
Answer :-
The possible values of a for the given problem are -1 and 5
Check :-
Case -1 :-
If a = -1 then ,the distance between A and B
=> AB = √[(2+1)²+(-2-2)²] units
=> AB = √[3²+(-4)²]
=> AB = √(9+16)
=> AB = √25
=> AB = ± 5 units
Therefore, AB = 5 units
Since, Distance can't be negative.
Case -2:-
If a = 5 then ,the distance between A and B
=> AB = √[(2-5)²+(-2-2)²] units
=> AB = √[(-3)²+(-4)²]
=> AB = √(9+16)
=> AB = √25
=> AB = ± 5 units
Therefore, AB = 5 units