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Question
Find the points on the x-axis which are at a distance of `2sqrt(5)` from the point (7, – 4). How many such points are there?
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Solution
We know that, every point on the x-axis in the form (x, 0).
Let P(x, 0) the point on the x-axis have `2sqrt(5)` distance from the point Q(7, – 4).
By given condition,
PQ = `2sqrt(5)` ...`[∵ "Distance formula" = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)]`
⇒ (PQ)2 = 4 × 5
⇒ (x – 7)2 + (0 + 4)2 = 20
⇒ x2 + 49 – 14x + 16 = 20
⇒ x2 – 14x + 65 – 20 = 0
⇒ x2 – 14x + 45 = 0
⇒ x2 – 9x – 5x + 45 = 0 ...[By factorisation method]
⇒ x(x – 9) – 5(x – 9) = 0
⇒ (x – 9)(x – 5) = 0
∴ x = 5, 9
Hence, there are two points lies on the axis, which are (5, 0) and (9, 0), have `2sqrt(5)` distance from the point (7, – 4).
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