# What point on the y-axis is 12sqrt13/13 units from the line defined by equation 3x -2y + 6 = 0?Answer: (0,-3) (0,9)

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A point on the on the y axis has an x-coefficient equal to 0. Let the point on the y-axis which is at a distance of 12/sqrt 13 from 3x - 2y + 6 = 0 be (0, y).

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

Substituting the value we have here:

12/ sqrt 13 = |a*0 -2*y + 6|/sqrt ( 3^2 + 4^2)

=>12/ sqrt 13 = |-2*y + 6|/sqrt ( 9 + 4)

=>12/ sqrt 13 = |-2*y + 6|/sqrt 13

=>|-2y + 6| = 12

=> -2y + 6 = 12 and -2y + 6 = -12

=> -2y = 6 and -2y = -18

=> y = 6/(-2) and y = -18/(-2)

=> y = -3 and y = 9

**The required point is (0,-3) and (0, 9)**

Given the equation of the line:

3x - 2y + 6 = 0

We need to find the point ( x,y) such that the distance between the line and the point is 12sqrt13

We will use thedistance between a point and a line.

D = l ax + by + c l / sqrt(a^2 + b^2)

==> 12sqrt13/ 13 = l 3x -2y + 6 l / sqrt(9+4)

==> 12sqrt13 / 13 = l 3x-2y + 6 l / sqrt13

Multiply and divide the right side by sqrt13.

==> 12sqrt13 = l 3x-2y + 6l sqrt13/ 13

==> l 3x -2y + 6 l = 12

==> But the point is on the y-axis, then x= 0

==> l -2y + 6 l = 12

Then, there are two cases:

1) ==> -2y + 6 = 12 ==> -2y = 6 ==> y= -3

2) ==> -(-2y+6) = 12 ==> 2y = 18==> y= 9

**Then the points are ( 0, -3) and (0, 9) .**