Find the point
on the x-axis which is equidistand from
(2,-5) and (-2, 9)
Answers
Step-by-step explanation:
Given :-
Two points (2,-5) and (-2, 9)
To find :-
Find the point on the x-axis which is equidistand from (2,-5) and (-2, 9)
Solution :-
Given points are :(2,-5) and (-2, 9)
Let A = (2,-5) and B = (-2,9)
Let the required point on the x-axis be P (x,0)
Since the equation of x-axis is y = 0
P be the equidistance from the given points A and B
=> PA = PB
On squaring both sides then
=> PA² = PB² -----------------(1)
Finding PA :-
Let (x1, y1)=P(x,0) => x1 = x and y1 = 0
Let (x2, y2)=A(2,-5)=>x2 = 2 and y2 = -5
We know that
The distance between two points (x1, y1) and
(x2, y2) is √[(x2-x1)²+(y2-y1)²] units
=> Distance between two points P and A
=> PA = √[(2-x)²+(-5-0)²]
=> PA =√[(2-x)²+(-5)²]
=> PA =√[(2-x)²+25]
On squaring both sides then
=> PA² = [√[(2-x)²+25]]²
=> PA² = (2-x)²+25
=> PA² = 4-4x+x²+25
=> PA² = x²-4x+29 -----------------(2)
And
Finding PB :-
Let (x1, y1)=P(x,0) => x1 = x and y1 = 0
Let (x2, y2)=B(-2,9)=>x2 = -2 and y2 = 9
We know that
The distance between two points (x1, y1) and
(x2, y2) is √[(x2-x1)²+(y2-y1)²] units
=> Distance between two points P and B
=> PB = √[(-2-x)²+(9-0)²]
=> PB=√[(-2-x)²+(9)²]
=> PA =√[(-(2+x))²+81]
On squaring both sides then
=> PB² = [√[(2+x)²+81]]²
=> PB² = (2+x)²+81
=> PB² = 4+4x+x²+81
=> PB² = x²+4x+85 -----------------(3)
From (1)
We have
(2) = (3)
=> PA² = PB²
=> x²-4x+29 = x²+4x+85
=> x²-4x+29-x²-4x-85 = 0
=> (x²-x²)+(-4x-4x)+(29-85) = 0
=> 0+(-8x)+(-56) = 0
=> -8x -56 = 0
=>-8x = 56
=> x = 56/-8
=> x = -56/8
=> x = -7
Therefore,the value of x = -7
The point P = (-7,0)
Answer:-
The required coordinates of the point P on the x-axis for the given problem is (-7,0)
Used formulae:-
-» The distance between two points (x1,y1) and (x2, y2) is √[(x2-x1)²+(y2-y1)²] units
-» Equation of x-axis is y=0