Perpendicular distance question?

Question

I've been having trouble with the following question: "The point $P(x,y)$ is equidistant from the lines $2x+y-3=0$ and $x-2y+1=0$, which intersect at $A$. Use the distance formula to show that $|2x+y-3|=|x-2y+1|$."

Would you have to initially find the point of intersection, then use the formula? Thanks.

Answer 1

Hints:

The distance from point $\;P=(a,b)\;$ to the line $\;Ax+By+C=0\;$ is given by

$$\frac{|Aa+Bb+C|}{\sqrt{A^2+B^2}}$$

It is a nice exercise, both in analytic geometry and/or in basic calculus, to prove the above indeed is the distance (i.e. the minimal distance) between the point and the line. Now solve your problem.

Answer 2

The signed distance of $(x_0,y_0)$ from $ax+by+c=0$ is $\displaystyle {{ax_0+by_0+c}\over{\sqrt{a^2+b^2}}}$.

In your problem you have two lines with $(a_1,b_1,c_1)=(2,1,-3)$ and $(a_2,b_2,c_2)=(1,-2,1)$. The denominator expression becomes the same. So you get your equation without any further step.