Find the point(s) on the x-axis which is at a distance of √41 units

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 (X₁, Y₁) and Q (X₂, Y₂) is given by:

PQ = $\sqrt {(\times_2 - \times _1)^2 + (Y_2 - Y_1)^2}$

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

∴ Distance AB = $\sqrt{(8 - \times)^2 + (- 5 - 0)^2}$

∴ AB = $\sqrt{\times^2 + 64 - 16 \times + 25}$

∴ AB = $\sqrt{\times ^2 - 16 \times + 89}$

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

∴ $\sqrt {\times ^2 - 16 \times + 89} = \sqrt 41$

∴ X² – 16 X + 89 = 41 (by squaring on both sides)

∴ X² – 16 X + 48 = 0

∴ X² – 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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