# Calculate the distance between the points (0,4) and the line 4x - 6y +12 = 0.

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We have the line:

4x-6y + 12 + 0

==> a= 4 b = -6 and c = 12

Also, we have the point (0, 4)

The formula for the distance between a line and a point is:

d = l ax + by + cl/sqrt(a^2 + b^2)\

= l 4*0 + -6*4 + 12 l / sqrt(16 + 36)

= l -12 l / sqrt(52)

= 12/2sqrt13 = 6/sqrt13

**Then the distance = 6/sqrt13**

We'll consider the fact that the equation of the line is given under the standard form:

ax + by + c = 0

We'll apply the formula for the distance from the given point to the given line, which is valid when the line equation is written in the standard form:

dist.D = |x*a + y*b + c|/sqrt(a^2 + b^2)

(x,y) - the coordinates of the given point

We'll identify a=4, b=-6 and c=12.

D = |0*4+4*(-6)+12|/sqrt[4^2+(-6)^2]

D = |-25+12|/sqrt(16+36)

D = 13/sqrt 52

D = 13/2sqrt13

D = 13*sqrt13/2*13

We'll reduce like terms:

**D = sqrt13/2**

So, the distance from the given point, to the line d is D = sqrt13/2.

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

Substituting the given values from the problem into

|x1*a + y1*b + c | / sqrt(a^2 + b^2), we get

|0*4 + 4*(-6) + 12 | / sqrt(4^2 + (-6)^2)

= |-24+12| / sqrt(16+36)

=12 / sqrt 52

**Therefore the required distance is 12 / sqrt 52**

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

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

The given point (x1,y1) = (0,4).

The line is 4x-6y+12 = 0.

So the distance d = |(4*0-6*4-12)/sqrt{(4^2+(-6)^2}

d = |-36/sqrt(16+36)|

d = 36/sqrt52

d = 18/sqrt13

d = (18*sqrt13)/13 is the distance of the point (0, 4)and the line 4x-6y+12 = 0.