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Question
Find the points on the x-axis, each of which is at a distance of 10 units from the point A(11, –8).
Solution
Let P(x, 0) be the point on the x-axis.
Then as per the question we have
AP = 10
`\implies sqrt((x - 11)^2 + (0 + 8)^2` = 10
`\implies` (x – 11)2 + 82 = 100 ...(Squaring both sides)
`\implies` ( x – 11)2 = 100 – 64
`\implies` ( x – 11)2 = 36
`\implies` x – 11 = ±6
`\implies` x - 11 = 6 or x - 11 = -6
`\implies` x = 6 + 11 or x = -6 + 11
`\implies` x = 17 or x = 5
Hence, the points on the x-axis are (17, 0) and (5, 0).
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The given points are collinear, so the area of the triangle formed by them is `square`.
∴ `1/2 |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = square`
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Hence, the relation between x and y is `square`.
