# Find if the distance between the point (2, y) and the line 3x-4y+5 = 0 is 12 units. Find y

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

Substituting the values we have here

D = |a*x1 + b*y1 + c|/sqrt (a^2 + b^2)

=> |3*2 - 4*y + 5|/sqrt (3^2 + 4^2)

=> |6 - 4y + 5|/ sqrt 25

Now D = 12

=> 60 = |11 - 4y|

=> 11 -4y = 60 and 11 - 4y = -60

=> 4y = -49 and -4y = -71

=> y = -49/4 and y = 71/ 4

**The values of y are y = -49/4 and y = 71/ 4**

We will use the distance between a line and a point formula to find y.

We know that :

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

Now we have D = 12, a = 3 b= -4 c = 5 x = 2 and y= y

Now we will substitute.

==> 12 = l 3*2 -4*y + 5 l / sqrt( 9+16)

=> 12 = l 11 -4yl / 5

Multiply by 5.

==> l 11-4y l = 60

Now we have 2 cases.

==> case(1) : (11-4y) = 60

==> -4y = 49 ==> y= -49/4

==> case(2) : -(11-4y) = 60

==> -11 + 4y = 60

==> 4y = 71

==> y= 71/4

**Then the values of y = { 71/4, -49/4}**