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Question
Find the points on the x-axis, whose distances from the `x/3 +y/4 = 1` are 4 units.
Solution
The given equation is: `"x"/3 + "y"/4 = 1`
multiplying by 12
4x + 3y – 12 = 0…………(i)
Suppose there is a point (a, 0) on the x-axis, then the distance of line (i) from point (a, 0) is
= `(4"a" + 0 - 12)/sqrt(16 + 9)`
= `± (4"a" - 12)/5`
∴ `± (4"a" - 12)/5 = 4`
or ± (4a – 12) = 20
By taking + ve sign 4a = 32 or a = 8
The required point on x-axis is (8, 0).
By taking – ve sign, `-(4"a" - 12)/5 = 4`
or –4a + 12 = 20
4a = –8, a = –2
The second required point is (–2, 0).
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