# Find the distance between the line y+2x -5= 0  and the point (1,3).

hala718 | Certified Educator

calendarEducator since 2008

starTop subjects are Math, Science, and Social Sciences

We have the line :

2x + y - 5 = 0

And the point (1,3)

The formula for the distance is:

d = l ax1 + by1 + cl / sqrt(a^2 + b^2)

= l 2 + 3 - 5l/ sqrt(4+1)

= 0/sqrt5=0

The sdistance is 0 units.

That means that the point (1,3) is actually on the line 2x+y -5 = 0

To check, ket us verify if the points on the lines by substitutting x and y va;ues:

2x+y - 5 = 0

2(1) + 3 -5 = 0

5-5=0

0=0

check Approved by eNotes Editorial

## Related Questions

krishna-agrawala | Student

All the four previous answers are correct and complete. However this question can also be solved in another way.

The given equation of line is:

y + 2x - 5 = 0

We substitute the values of coordinates of the given point (1, 3) in the left hand side of the equation and find its value. Thus we get

3 + 2*1 - 5

= 3 + 2 - 5

= 0

This is equal to the right hand side of the equation of line.

Therefore the given point lies on the line.

Therefore the distance between the line and the point = 0

giorgiana1976 | Student

The distance from a point to a given lineis the perpendicular to that line.

We have to determine the equation of the line that passes through the given points.

We'll note the given line by l:y+2x -5= 0

Because we don't know the coordinates of the intercepting point of 2 perpendicular lines, we'll calculate the distance using the formula:

d(point,l)=|(3+2-5)|/sqrt[(2)^2+(-5)^2]

d(point,l) = 0/sqrt 29

d(point,l) = 0

So, there is no distance between the point and the given line, meaning that the point is located on the line.

We'll prove that by substituting the coordinates of the point in the eq. of the line, y+2x -5= 0.

3 + 2*1 - 5 = 0

0 = 0 q.e.d

thewriter | Student

The distance between the line y=mx+b and (x1,y1) is given by |y1-mx1-b|/sqrt(m^2+1).

Now the line we have is y+2x-5=0 or y=-2x+5

The point is (1,3)

Substituting the values in the formula |y1-mx1-b|/sqrt(m^2+1) we get |3-(-2)*1-5|/sqrt((-2)^2+1)=|3+2-5|/sqrt(4+1)= 0

Therefore the point lies on the line and the required distance is 0.

neela | Student

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

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

Substitute for (x1,y1) by (1,3) and  for (ax+by+c = 0)  by  (y+2x-5 = 0) or (2x+y-5 = 0) and we get:

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

d =  |(2+3-5)/(5^(1/2)|

d = (5-5)/5^(1/2).

d = 0/sqrt5 = 0