Question

How far is the point (3,1) from the line 5x + 12y = 1?

Answer 1

Task. How far is the point $(3,1)$ from the line $5x+12y=1$?

Solution. To find the distance from a point $A(\bar{x},\bar{y})$ from the line

$ax+by+c=0$

we should substitute the coordinates of point $A$ into the normal equation of the line, i.e. the equation divided by $\sqrt{a^{2}+b^{2}}$:

$dist=\frac{a\bar{x}+b\bar{y}+c}{\sqrt{a^{2}+b^{2}}}.$

In our case the distance from point $(3,1)$ to the line $5x+12y-1=0$ is equal to

$dist=\frac{|5*3+12*1-1|}{\sqrt{5^{2}+12^{2}}}=\frac{26}{\sqrt{25+144}}=\frac{26}{\sqrt{169}}=\frac{26}{13}=2.$