Find the points on the line x + y = 1 that lie at a distance 3 units from the line 5x+12y = 3.

Answer

`(2-sqrt(2), -1+sqrt(2)) " and " (2+2sqrt(5), -1+sqrt(5))`

Explanation

Given lines intersect at P(2,-1).
Slope of line x+y-1 = 0 is -1.
`therefore "tan"theta = -1`
`therefore "cos"theta = -(1)/(sqrt(2)), "sin" theta = (1)/(sqrt(2))`
One particle moves 2 units upward from point P on the above line.
Thus coordinates of new position obtained by the particle are
`(2+2(-(1)/(sqrt(2))), -1+2 * (1)/(sqrt(2)))-= (2-sqrt(2), -1+sqrt(2))`
Slope of line x-2y-4=0 is 1/2.
`therefore " tan" theta = (1)/(2)`
`therefore " cos" theta = (2)/(sqrt(5)), "sin" theta = (1)/(sqrt(5))`
Other particle moves 5 units upward from point P on above line.
Then coordinates of new position obtained by the particle are
`(2+5((2)/(sqrt(5))), -1+5 * (1)/(sqrt(5)))-= (2+2sqrt(5), -1+sqrt(5))`