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

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

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The distance between a line ax+by+c = 0 and the point (x1, y1) is:

|a*x1 + b*y1 + c|/sqrt (a^2 + b^2)

Here the line is x - y + 3 = 0 and the point is (2m , m-1). The distance between them is 2

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

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

=> m+ 4 = 2*sqt 2 and m + 4 = -2*sqrt 2

=> m = 2*sqrt 2 - 4 and m = -2*sqrt 2 - 4

The values that m can take are 2*sqrt 2 - 4 and -2*sqrt 2 - 4

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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)}.

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