Find a locus of points to satisfy these conditions?

Question

So, we have two straight lines: $$3x-4y+5=0$$ $$2x+3y-4=0$$ You have to find a locus of points from which all perpendiculars to the two lines given are in a 2:3 ratio.

Answer 1

Let $\,(a,b)\in\Bbb R^2\,$ be such a point. Its distances from the given lines are

$$\frac{|3a-4b+5|}{\sqrt{3^2+4^2}}=\frac{1}{5}|3a-4b+5|$$

$$\frac{|2a+3b-4|}{\sqrt{2^2+3^2}}=\frac{1}{\sqrt{13}}|2a+3b-4|$$

Thus, for the wanted ratio to be what we want, it must be that

$$\frac{3}{5}|3a-4b+5|=\frac{2}{\sqrt{13}}|2a+3b-4|$$

You can either check signs and solve different cases, or you can also square both sides, say

$$\frac{9}{25}(9a^2-24ab+16b^2+30a-40b+25)=\frac{4}{13}(4a^2+12ab+9b^2-16a-24b+16)\iff$$

$$\left(\frac{81}{25}-\frac{16}{13}\right)a^2+\left(\frac{144}{25}-\frac{36}{13}\right)b^2-\left(\frac{216}{25}+\frac{48}{13}\right)ab+\ldots \,\text{etc.}$$

Yes, it looks really ugly, but after all it's only a quadratic in two variables. If you already know how to classify them by means of matrices then nice, otherwise...tough world.