Given the distance from the point 2m m1 to the line xy30 d2

Given the distance from the point (2m, m-1) to the line x-y+3=0, d=2 find m.

Solution

We\'ll recall the identity that represents the distance from a point to a line:

d = |a*xP + b*yP + c|/sqrt(a^2 + b^2)

the line is: ax + by + c = 0 and the coordinates of the point are: P(xP,yP).

Comparing, we\'ll get:

2 = |1*2m - 1*(m-1) + 3|/sqrt(1^2 + 1^2)

2 = |2m - m + 1+ 3|/sqrt 2

2sqrt2 = |m + 4|

|m + 4| = 2sqrt2

We\'ll discuss 2 cases:

1) m + 4 = 2sqrt2

m = 2sqrt2 - 4

m = 2(sqrt2 - 2)

2) m + 4 = -2sqrt2

m = -4 - 2sqrt2

m = -2(sqrt2 + 2)

If the distance from the point (2m,m-1) to the line x-y+3 = 0 is d=2, the values of m are: {-2(sqrt2 + 2) ; 2(sqrt2 - 2)}.