# Real numbers.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

The distance between a point (x1 , y1) and the line ax + by + c = 0 is given by |a*x1 + b*y1 + c|/sqrt (a^2 + b^2)

The line we have 3x-4y-1 = 0 and the distance from the point (m, m+ 1) is 1

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

=> 5 = |3m - 4m - 4 - 1|

=> 5 = |-m - 5|

=> -m- 5 = 5

=> m = -10

and -m-5 = -5

=> m = 0

**The real number m = 0 and 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).**