8. Find the values of x for which the distance between the point A(2,-3) and Q(x,5) is 10.
Answers
Step-by-step explanation:
Given :-
The distance between the point A(2,-3) and Q(x,5) is 10.
To find :-
Find the values of x ?
Solution :-
Given points are A(2,-3) and Q(x,5)
Let (x1, y1) = A(2,-3) => x1 = 2 and y1 = -3
Let (x2, y2) = (x,5) => x2 = x and y2 = 5
We know that
The distance between the two points (x1, y1) and (x2, y2) is √[(x2-x1)²+(y2-y1)²] units
The distance between the two points A and Q
=> AQ = √[(x-2)²+(5-(-3))²]
=> AQ = √[(x-2)²+(5+3)²]
=> AQ = √[(x-2)²+8²]
=> AQ = √(x²-4x+4+64)
=> AQ = √(x²-4x+68) units
According to the given problem
The distance between A and Q = 10 units
=> √(x²-4x+68) = 10
On squaring both sides then
=> [√(x²-4x+68)]² = 10²
=> x²-4x+68 = 100
=> x²-4x+68-100 = 0
=> x²-4x-32 = 0
=> x²-8x+4x-32 = 0
=> x(x-8)+4(x-8) = 0
=> (x-8)(x+4) = 0
=> x-8 = 0 (or) x+4 = 0
=> x = 8 (or) x = -4
Therefore, x = 8 and -4
Answer:-
The values of x for the given problem are 8 and -4
Used formulae:-
Distance Formula :-
The distance between the two points (x1, y1) and (x2, y2) is √[(x2-x1)²+(y2-y1)²] units
Used Identity :-
→ (a-b)² = a²-2ab+b²