find the point on the x axis which is equidistant from (3, -6) and (-3, 10)
Answers
Answer:If x and y are two independent binomial variables with parameters 6 and ½ and 4 and ½ respectively, what isP(x +y ≥ 1)?
Step-by-step explanation:
Step-by-step explanation:
Given :-
The points are (3, -6) and (-3, 10)
To find :-
Find the point on the x axis which is equidistant from (3, -6) and (-3, 10) ?
Solution :-
Given points are (3, -6) and (-3, 10)
Let A = (3,-6)
Let B = (-3,10)
Let the point on the x -axis which is equidistant from the given points be P (x,0)
Since, The equation of x -axis is y = 0
According to the given problem
AP = PB
Finding AP :-
Let (x1, y1) = A( 3,-6 ) => x1 = 3 and y1 = -6
Let (x2, y2) = P(x,0) => x2 = x and y2 = 0
We know that
The distance between (x1, y1) and (x2, y2) is
√[(x2-x1)²+(y2-y1)²] units
=> AP = √[(x-3)²+(0-(-6))²]
=> AP = √[(x-3)²+(0+6)²]
=> AP = √(x-3)²+(6)²]
=> AP = √(x²-6x+9+36)
=> AP = √(x²-6x+45) units ----------(1)
Finding PB :-
Let (x1, y1) = P( x,0 ) => x1 = x and y1 = 0
Let (x2, y2) = B(-3,10) => x2 = -3 and y2 = 10
We know that
The distance between (x1, y1) and (x2, y2) is
√[(x2-x1)²+(y2-y1)²] units
=> PB = √[(-3-x)²+(10-0)²]
=> PB = √[(-(3+x))²+(10)²]
=> PB = √(x²+6x+9+100)
=> PB = √(x²+6x+109) units ----------(2)
Now,
AP = PB
=> √(x²-6x+45) = √(x²+6x+109)
On squaring both sides then
=> [√(x²-6x+45)]²= [√(x²+6x+109)]²
=> x²-6x+45 = x²+6x+109
=> -6x+45 = 6x+109
=> -6x-6x = 109-45
=> -12x = 64
=> x = -64/12
=> x = -16/3
The value of x = -16/3
The point P = (-16/3 ,0)
Answer:-
The required point for the given problem is (-16/3 ,0)
Used formulae:-
→ The distance between (x1, y1) and (x2, y2) is
√[(x2-x1)²+(y2-y1)²] units
→ The equation of x -axis is y = 0
