**Q) Find the point(s) on the x-axis which is at a distance of √41 units from the point (8, -5).**

**Ans: **

Let the point be A and coordinates of this point be (x, y)

Since any point on X-axis will have y = 0,

Hence, the coordinates of the point A will be (X, 0)

We know that the distance between two points P (X_{1}, Y_{1}) and Q (X_{2}, Y_{2}) is given by:

PQ =

We have coordinates as A (X,0) and B (8, -5)

∴ Distance AB =

∴ AB =

∴ AB =

∵ It is given that distance between these 2 points is √41

∴

∴ X^{2} – 16 X + 89 = 41 (by squaring on both sides)

∴ X^{2} – 16 X + 48 = 0

∴ X^{2} – 12 X – 4 X + 48 = 0

∴ X (X – 12) – 4 (X – 12) = 0

∴ (X – 4) (X – 12) = 0

∴ X = 4 and X = 12

**Since we get two values of X, it means there will be two points and the coordinates of these two points will be (4, 0) and (12, 0)**

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