Q.Find the point on x-axis which is equidistant from the points (7, -6) and (3, 4).

Three distance formulas underpin this chapter's metric questions:

1. Point to point. $D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ (the ordinary Pythagorean distance).
2. Point to line. The distance from $P(x_1,y_1)$ to the line $ax+by+c=0$ is

proved by comparing the normal form of the given line with the normal form of the parallel line through $P$, and taking the difference of their normal lengths from the origin.

3. Line to parallel line. For $a_1x+b_1y+c_1=0$ and $a_1x+b_1y+c_2=0$ (same $a_1,b_1$), $D=\dfrac{|c_2-c_1|}{\sqrt{a_1^2+b_1^2}}$ — a special case of formula 2, taking the point as the origin (or any point) on one of the two lines.

Foot of the perpendicular and image of a point. From $P(x_1,y_1)$, the foot of the perpendicular on $ax+by+c=0$ is found from the parametric relation

and the image (reflection) of $P$ in that line doubles the same offset:

(The foot is the midpoint of $P$ and its image.) The same idea — reflect a point in a line, most often the $x$-axis or $y$-axis — is the standard tool for 'ray of light reflected off a surface' problems: the angle of incidence equals the angle of reflection exactly when the reflected ray is the straight line joining the image of the source to the target point.

Position of a point relative to a line and the acute-angle test. A point $P(x_1,y_1)$ lies on the origin side or non-origin side of $ax+by+c=0$ ($c\ne0$) according as $ax_1+by_1+c$ has the same or opposite sign as $c$. …