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Question
If sum of the perpendicular distances of a variable point P (x, y) from the lines x + y – 5 = 0 and 3x – 2y+ 7 = 0 is always 10. Show that P must move on a line.
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Solution
The equations of the given lines are
x + y – 5 = 0 … (1)
3x – 2y + 7 = 0 … (2)
The perpendicular distances of P (x, y) from lines (1) and (2) are respectively given by
`d_1 = |x + y - 5|/(sqrt((1)^2 + (1)^2)` and `d_2 = |3x - 2y + 7|/(sqrt((3)^2 + (2)^2)`
i.e., `d_1 = (x + y - 5)/sqrt2` and `d_2 = |3x -2y + 7|/sqrt(13)`
It is given that d1 + d2 = 10
`= (x + y - 5)/sqrt2 + |3x -2y + 7|/sqrt(13) = 10`
= `sqrt13 |x + y - 5| + sqrt2 |3x -2y + 7|-10sqrt26 = 0`
= `sqrt13 |x + y - 5| + sqrt2 |3x -2y + 7|-10sqrt26 = 0`
[Assuming (x + y - 5) and (3x - 2y + 7) are positive]
= `sqrt13x + sqrt13y - 5sqrt13 + 3sqrt2x - 2sqrt2y + 7sqrt2 - 10sqrt26 = 0`
= `x(sqrt13x + 3sqrt2) + y (sqrt13 - 2sqrt2) + (7sqrt2 - 5sqrt13 - 10sqrt26) = 0` which is the equation of a line.
Similarly, we can obtain the equation of line for any signs of (x + y -5) and (3x - 2y + 7)
Thus, point P must move on a line.
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