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Question
What are the points on the y-axis whose distance from the line `x/3 + y/4 = 1` is 4 units.
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Solution
Let (0, b) be the point on the y-axis whose distance from line `x/3 + y/4 = 1` is 4 units.
The given line can be written as 4x + 3y – 12 = 0 ...(1)
On comparing equation (1) to the general equation of line Ax + By + C = 0, we obtain A = 4, B = 3, and C= –12.
It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by
`d = |Ax_1 + By_1 + C|/sqrt(A^2 + B^2)`
Therefore, if (0, b) is the point on the y-axis whose distance from line `x/3 + y/4 = 1` is 4 units, then:
`4 = |4(0) + 3(b) -12|/sqrt(4^2 + 3^2)`
4 = `|3b - 12|/5`
= 20 = |3b - 12|
= 20 = ± (3b - 12)
= 20 = (3b - 12) or 20 = -(3b - 12)
= 3b = 20 + 12 or 3b = -20 + 12
= `b = 32/3` or `b = 8/3`
Thus, the required points are `0, 32/3` and `(0, 8/3)`.
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