Find a point on the x-axis which is equidistant from the points (7, 6) and (-3, 4).
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Answer:
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Given : points (7, 6) and (-3, 4).
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)
Let A (7, 6) , B(-3, 4) be the points.
Given :A (7, 6) , B(-3, 4) are equidistant from P( x,0), AP = BP .
For distance AP : A (7, 6) , P( x,0)
x1 = 7 ,y1 = 6 , x2 = x & y2 = 0
Distance of AP =√ (x − 7)² + (0 − 6)²
[Distance Formula =√(x2 - x1)² + (y2 - y1)²]
AP = √( x - 7)² + (6)²
= √[x² + 7² - 2 × x × x 7 + 6²]
[(a - b)² = a² + b² - 2ab]
AP = √x² + 49 - 14x + 36
AP = √x² - 14 x + 85…………………(1)
For distance BP : B(-3, 4), P( x,0)
x1 = -3 , x2 = x & y1 = 4 , y2 = 0
BP = √ (x - (-3))² + ( 0 - 4)²
[Distance Formula = √(x2 - x1)² + (y2 - y1)²]
BP = √(x + 3 )² + ( -4 )²
BP = √x² + 9 + 6 x + 16
[(a + b)² = a² + b² + 2ab]
BP = √x² + 6x + 25…………………..(2)
AP = BP [equidistant from P]
√x² - 14 x + 85 = √x² + 6x + 25
[From eq. 1 and 2 ]
On squaring both sides
x² - 14 x + 85 = x² + 6 x + 25
x² - x² - 14 x - 6x = 25 - 85
-20 x = -60
x = 60/20
x = 3
Hence , the point on x - axis is (3 , 0) .
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