Question 859827
<pre>
The perpendicular distance from the point 

(x<sub>1</sub>,y<sub>1</sub>)

to the line 

Ax + By + C = 0 is

d = {{{abs((Ax[1]+By[1]+C)/sqrt(A^2+B^2))}}}
 
In your case, (x<sub>1</sub>,y<sub>1</sub>) = (-6,-5)

And the line is

 y = -2x - 2.

but we must first get it in the form

Ax + By + C = 0

 y = -2x - 2

2x + y + 2 = 0

So A = 2, B = 1, C = 2, x<sub>1</sub> = -6, y<sub>1</sub> = -5.

d = {{{abs((Ax[1]+By[1]+C)/sqrt(A^2+B^2))}}}


d = {{{abs((2(-6)+1(-5)+2)/sqrt(2^2+1^2))}}}

d = {{{abs((-12-5+2)/sqrt(4+1))}}}

d = {{{abs((-15)/sqrt(5))}}}

d = {{{(15)/sqrt(5)}}}

Rationalize the denominator:

d = {{{(15/sqrt(5))(sqrt(5)/sqrt(5))}}}

d = {{{(15sqrt(5))/5}}}

d = {{{3sqrt(5)}}}

Edwin</pre>