Find the distance between the point (2,-5) and the line 3x-5y+13= 0

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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)

Here the point is (2, -5) and the line is 3x - 5y + 13 = 0

The distance D = |3*2 - 5*(-5) + 13|/ sqrt ( 9 + 25)

=> |6 + 25 + 13|/sqrt 34

=> 44/sqrt 34

The distance between the point (2,-5) and the line 3x-5y+13= 0 is 44/ sqrt 34

hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

For the distance between a point and a line we use the formula:

D = l ax1 + by1 + c l/ sqrt(a^2+b^2)

Such that: ax+by+c = 0 is the line and (x1,y1) is the point.

In our example, we have the line 3x-5y + 13 = 0

==> a =3    b= -5    c = 13

and the point (2,-5) ==> x1= 2   y1= -5

Let us substitute:

==> D= l 3*2 + -5*-5 + 13 l / sqrt(9+25)

           = l 6+25 + 13l / sqrt(34)

          = 44/sqrt34

Then the distance between the point and the line is 44/sqrt34.

 

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