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प्रश्न
Find the co-ordinates of points on the x-axis which are at a distance of 17 units from the point (11, -8).
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उत्तर
Let the coordinates of the point on x-axis be (x, 0).
From the given information, we have:
`sqrt((x -11)^2 + (0 + 8)^2)` = 17
(x - 11)2 + (0 + 8)2 = 289
x2 + 121 - 22x + 64 = 289
x2 - 22x - 104 = 0
x2 - 26x + 4x - 104 = 0
x(x - 26) + 4(x - 26) = 0
(x - 26)(x + 4) = 0
x = 26, -4
Thus, the required co-ordinates of the points on x-axis are (26, 0) and (-4, 0).
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By distance formula,
PQ = `sqrt(square + (y_2 - y_1)^2`
= `sqrt(square + square)`
= `sqrt(square + square)`
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= `sqrt(125)`
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