Find the point on x-axis which is equidistant from the points (-2, 5) and (2,-3).
Answers
Answer:
point on x axis is (-2,0)
Step-by-step explanation:
let the point on x-axis be x(x,0)
GIVEN:A(-2,5) and B(2 ,-3) are equidistant from p,so AX =BX then,
∴AX =BX = AX²=BX²
∴(AX)²= (X+2)² + 25
∴(BX)²= (X-2)² = 9
∴X² + 4x + 4 + 25 = X² - 4X + 4 +9
∴8X = - 16
∴X=-2
HENCE, POINT ON X AXIS IS -2.0
Given : points (-2, 5) and (2,-3)
To prove : a point on the x-axis which is equidistant
Solution :
The coordinates of every point on the x- axis are of the form (x,0)
Given : A( - 2 ,5) and B( 2 , - 3 ) are equidistant from P( x,0), AP = BP .
For distance AP : x1 = - 2 ,y1 = 5 , x2 = x & y2 = 0
Distance of AP =√ (x- (-2)² + (0 - 5)²
[Distance Formula=√(x2 - x1)² + (y2 - y1)²]
AP = √( x + 2 )² + (- 5 )²
= √x² + 2² + 2 × 2 × x + 5²
[(a + b)² = a² + b² + 2ab]
AP = √x² +4 +4x +25
AP = √x² + 4 x + 29…………………(1)
For distance BP : x1 = 2 , x2 = x & y1 = - 3 , y2 = 0
BP =√ (x −2) ² + ( 0 −(−3))²
[Distance Formula = √(x2 - x1)² + (y2 - y1)²]
BP = √(x - 2 )² + ( 3 )²
BP= √x² + 4 - 4 x + 9
[(a- b)² = a² + b² - 2ab]
BP= √x² - 4 x + 13…………………..(2)
AP = BP [equidistant from P]
√x² + 4 x + 29 = √x² - 4 x + 13
[From eq. 1 and 2 ]
On squaring both sides
x² + 4 x + 29 = x² - 4 x + 13
x² - x² + 4 x + 4 x= + 13 - 29
8 x = - 16
x = -16/8
x = - 2
Hence , the point on x - axis is (-2 , 0) .
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