Find a point on y-axis which is equidistant from the points (5, - 2) and (- 3, 2).
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Given : points (5, - 2) and (- 3, 2)..
To prove : a point on the y-axis which is equidistant
Solution :
The coordinates of every point on the y- axis are of the form (0,y)
Let A (7, 6) , B(-3, 4) be the points.
Given : A (7, 6) , B(-3, 4) are equidistant from P( x,0), PA = PB
⇒ PA² = PB²
By using Distance Formula =√(x2 - x1)² + (y2 - y1)²:
(0 - (- 5))² + (y - (- 2))² = (0 - 3)² + (y - 2)²
(5)² + (y + 2)² = (- 3)² + (y - 2)²
25 + y² + 4y + 4 = 9 + y² - 4y + 4
[(a ± b)² = a² + b² ± 2ab]
29 + y² + 4y = 13 + y² - 4y
4y + 4y = 13 - 29
8y = - 16
y = - 16/8
y = - 2
Hence , the point on y - axis is (0 ,- 2) .
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