# Thread: Find the distance of the point

1. ## Find the distance of the point

Find the distance of the point (-6,-6) from the line through (9,5) which points in the direction of -1I-2J.

2. The distance the point $(p,q)$ is from the line $Ax+By+C=0$ equals $\frac{{\left| {Ap + Bq + C} \right|}}{{\sqrt {A^2 + B^2 } }}$.

3. Originally Posted by Plato
The distance the point $(p,q)$ is from the line $Ax+By+C=0$ equals $\frac{{\left| {Ap + Bq + C} \right|}}{{\sqrt {A^2 + B^2 } }}$.
so (p,q) is (-6,-6), how do I figure out $Ax+By+C=0$?

4. Originally Posted by viet
so (p,q) is (-6,-6), how do I figure out $Ax+By+C=0$?
That is a problem you need to solve.
How does one write the equation of a line given a point and a direction vector?

5. using the parametric equations it should be: x = 9-1t, y= 5-2t

6. Originally Posted by viet
using the parametric equations it should be: x = 9-1t, y= 5-2t
Now write it in the Ax+By+C =0 form.