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let L: 4x + 3y + C = 0 be a line of unit distance and parallel to the line 4x + 3y = 10
then (C + 10)/sqrt(4^2 + 3^2) = +/- 1
then C = -5 or -15
then L: 4x + 3y - 5 = 0 ---(1)
or 4x + 3y - 15 = 0 ---(2)
the intersecting point of x + y = 4 ---(3)with L is the required point
4(3) - (1), we get y = 11 ==> x = -7 ie the point is (-7, 11)
or 4(3) - (2), we get y = 1 ==> x = 3 ie the point is (3, 1)
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