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

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

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

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

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.