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

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### 2 Answers

The distance between the point (x1 , y1) and the 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 = 5 = |4*2 + 3*(-3) - C|/sqrt (16 + 9)

=> 5 = |8 - 9 - C| / 5

=> 25 = |-1 - C|

-25 = 1 + C

=> C = -26

Also 25 = -1 - C

=> C = 24

**The values of C are C = -26 and C= 24**

We will use the distance between a point an a line formula to find the value of c.

We know that:

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

We will substitute:

==> 5 = l 4*2 + 3*-3 + -C l / sqrt(4^2+3^2)

==> 5= l -1-c l / 5

==> l -1-cl = 25

==> -1-c = 25 ==> c = -26

==> 1+c = 25 ==> c = 24

Then we have two possible value for c.

**==> C = { -26, 24}**