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Question
Answer the following question:
Find points on the X-axis whose distance from the line `x/3 + y/4` = 1 is 4 unit
Solution
Let P(a, 0) be the required point on the X-axis.
Now, distance of P from the line `x/3 + y/4` = 1,
i.e., 4x + 3y – 12 = 0 is 4 units.
∴ `|(4"a" + 3(0) - 12)/sqrt(4^2 + 3^2)|` = 4
∴ `|(4"a" - 12)/5|` = 4
∴ 4a – 12 = ± 20
∴ 4a – 12 = 20 or 4a – 12 = – 20
∴ 4a = 32 or 4a = – 8
∴ a = 8 or a = – 2
Hence, coordinates of required points on X-axis are (8, 0) and (– 2, 0).
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