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Let $M(1,-1)$ be a point in a plane. Find its distance from a line given by $x+2y-4=0$.

Later on I found a formula: $$d=\frac{\left | Ax_{0}+Bx_{0}+C \right | }{\sqrt{A^2+B^2}}$$

But I did it somehow without using it. Like this:

The distance between a point $M$ and a line $p:\; y=-\frac{1}{2}x+2$ is a segment of another line that is perpendicular to $p$, let's mark it as $q$, which has the property that $M\in q$.
From $M\in q$ we have: $$y+1=k(x-1) \Rightarrow q:\;\; y=kx-(k+1)$$ From $p\perp q$ we have: $$k_{1}k_{2}=-1 \; \Rightarrow \; -\frac{1}{2}k_{2}=-1 \; \Rightarrow \; k_{2}=2$$ $$q: \;\; y=2x-3$$
Now, we obtain a point of intersection $N$ between $p$ and $q$ which is $N(2,1)$.
And now we find the distance between points $M(1,-1)$ and $N$. $$d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}=\sqrt{1^2+2^2} \;\; \Rightarrow \;\; d=\sqrt{5}$$

tyr
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1 Answers1

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This is just an answer to get your question formally answered.

Yes, it is.

Since @Valerin has already confirmed that you have answered it "in a nice way" [in the body of the question itself], I just add to @colormegone's comment ("If you apply your method for a general line and external point, you will now know one way to derive the formula you mention.") the following general method (copied from this answer of mine):

The distance $d$ between a point $P_1(x_{1},y_{1})$ and a line $r$ whose equation is $Ax+By+C=0$ can be derived algebraically as follows:

i) Find the equation of the straight line $s$ passing through $P_1$ and which is orthogonal to $r$. Call $P_2$ the intersecting point of $r$ and $s$.

ii) Find the co-ordinates of $P_2(x_{2},y_{2})$.

iii) Find the distance from $P_1$ to $P_2$. This distance is $d$.

Américo Tavares
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