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Question
Find a point which is equidistant from the points A(–5, 4) and B(–1, 6)? How many such points are there?
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Solution
Let P(h, k) be the point which is equidistant from the points A(–5, 4) and B(–1, 6).
∴ PA = PB ...`[∵ "By distance formula, distance" = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)]`
⇒ (PA)2 = (PB)2
⇒ (– 5 – h)2 + (4 – k)2 = (– 1 – h)2 + (6 – k)2
⇒ 25 + h2 + 10h + 16 + k2 – 8k = 1 + h2 + 2h + 36 + k2 – 12k
⇒ 25 + 10h + 16 – 8k = 1 + 2h + 36 – 12k
⇒ 8h + 4k + 41 – 37 = 0
⇒ 8h + 4k + 4 = 0
⇒ 2h + k + 1 = 0 ...(i)
Mid-point of AB = `((-5 - 1)/2, (4 + 6)/2)` = (– 3, 5) ...`[∵ "Mid-point" = ((x_1 + x_2)/2, (y_1 + y_2)/2)]`
At point (– 3, 5), from equation (i),
2h + k = 2(– 3) + 5
= – 6 + 5
= – 1
⇒ 2h + k + 1 = 0
So, the mid-point of AB satisfy the equation (i).
Hence, infinite number of points, in fact all points which are solution of the equation 2h + k + 1 = 0, are equidistant from the points A and B.
Replacing h, k by x, y in above equation, we have 2x + y + 1 = 0
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