# Find the distance between the line 3x+4y=11 and the point (2,5).

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### 2 Answers

We are given the coordinates of the point as ( 2 , 5) and the equation of the line is 3x + 4y = 11 or 3x + 4y - 11 = 0.

Now the relation for calculating the distance d of a point (x1, y1) from the line ax+by +c = 0, is:

d = |ax1+by1+c|/ sqrt (a^2+b^2)

Substituting the values we have

d = | 3*2 + 5* 4 - 11| / sqrt ( 3^2 + 4^2)

=> d = | 6 + 20 -11| / sqrt 25

=> d = 15 / 5

=> d = 3

**Therefore the required distance is 3.**

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

d = |(ax1+by1+c)/(a^2+b2)^(1/2)|.

Therefore the distance between the the line 3x+4y = 11, Or 3x+4y-11 = 0 and the point (2,-3) is given by:

d = |(3*2+4*5-11)/(3^2+4^2)^(1/2)|

d = |(6+20-11)/(25)^(1/2)|

d = 15/5

d= 3.

**So the distance between 3x+4y= 11 and (2,5) is 3 units.**