# Find the real number m if the distance from the point(m, m+1) to the line 3x-4y-1=0 is 1.

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### 2 Answers

We have to find the real number m if the distance from (m, m+1) to the line 3x-4y-1=0 is 1.

Now the distance between a point and a line can be given by the relation D = |a*x1 + b*y1 + c|/ sqrt ( a^2 + b^2)

Here a = 3, b = -4 and c = -1, x1 = m and y1 = (m+1)

=> 1 = |3*m -4*(m+1) -1|/ sqrt ( 3^2 + (-4)^2)

=> 1 = (3m - 4m - 4 - 1)/ 5

=> -m - 5 = 5

=> m = -10.

**Therefore m = -10.**

We'll note the given point as A(m, m+1).

We'll write the formula of the distance from a given point to a given line:

d(A, line): |3*m - 4*(m+1) - 1|/sqrt[3^2 + (-4)^2]

But d(A, line) = 1

|3*m - 4*(m+1) - 1|/sqrt[3^2 + (-4)^2] = 1

(3*m - 4*(m+1) - 1)/sqrt(9+16) = 1

We'll multiply by sqrt 25 both sides:

3m - 4m - 4 - 1 = sqrt25

We'll combine like terms:

-m - 5 = 5

We'll add 5 both sides:

-m = 10

m = -10

**The coordinates of the point A are: A(-10 , -9).**