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Question
Find the point on the X–axis which is equidistant from A(–3, 4) and B(1, –4).
Solution
Let point C be on the X-axis which is equidistant from points A and B.
Point C lies on X-axis.
∴ Its y co-ordinate is 0.
Let C = (x, 0)
C is equidistant from points A and B.
∴ AC = BC
According to distance formula,
`∴ sqrt([x – (– 3)]^2 + (0 – 4)^2] = sqrt((x – 1)^2 + [0 – ( – 4)]^2)`
∴ [x – (– 3)]2 + (0 – 4)2 = (x – 1)2 + [0 – ( – 4)]2 ...(Squaring both the sides)
∴ (x + 3)2 + (– 4)2 = (x – 1)2 + 42
∴ x2 + 6x + 9 + 16 = x2 – 2x + 1 + 16
∴ `cancelx^2 + 6x + 25 = cancelx^2 – 2x + 17`
∴ 6x + 2x = 17 – 25
∴ 8x = – 8
∴ `"x" = (-8)/8`
∴ x = –1
∴ The point on the X-axis, which is equidistant from points A and B, is (–1, 0).
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