Question:
find points on the y axis which are at a distance of 10 units from the point 8,8
Answer:
Let P(0, y) be the required point on y-axis.
Given, distance between the points (0, y) and (8, 8) is 10
=> √{(8 - 0)2 + (8 - y)2 } = 10
=> √{64 + (8 - y)2 } = 10
Squaring on both side, we get
64 + (8 - y)2 = (10)2
=> 64 + 64 + y2 - 16y = 100
=> 128 + y2 - 16y = 100
=> y2 - 16y + 128 - 100 = 0
=> y2 - 16y + 28 = 0
=> y2 - 14y - 2x + 28 = 0
=> y(y - 14) - 2(x - 14) = 0
=> ( y - 14) * (y - 2) = 0
=> y = 2, 14
So, the required point on y-axis is (0, 2) and (0, 14)
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