# Thread: Distance from a point to a line

1. ## Distance from a point to a line

Use the distance from a point to a line formula to check the distance from P(7, -4) to 3x - y =5

okay the slope of the line is -3/1, which is perpendicular to 1/3

so the equation i think is y+4/x-7
which ive then written 3(y+4)=1(x-7)
then 3y+12 = x-7
and i wrote x +3y = 19,

i'm not sure if what i just wrote was correct, and i don't know where to go from there..

2. ## Re: Distance from a point to a line

Well for starters, your line is \displaystyle \begin{align*} 3x - y = 5 \end{align*}, which becomes \displaystyle \begin{align*} y = 3x - 5 \end{align*}, and so its gradient is 3, not -3. Therefore the perpendicular line should have gradient \displaystyle \begin{align*} -\frac{1}{3} \end{align*}.

3. ## Re: Distance from a point to a line

There is a formula for the distance between the line ax+by+c=0 to the point (x1,y1)

It is
$\text{distance}=\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2 }}$

Before applying the formula make sure the equation of the line is in that form exactly.

4. ## Re: Distance from a point to a line

Originally Posted by Shakarri
There is a formula for the distance between the line ax+by+c=0 to the point (x1,y1)

It is
$\text{distance}=\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2 }}$

Before applying the formula make sure the equation of the line is in that form exactly.
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