Three distance formulas underpin this chapter's metric questions:
Point to point.D=(x2−x1)2+(y2−y1)2 (the ordinary Pythagorean distance).
Point to line. The distance from P(x1,y1) to the line ax+by+c=0 is
D=a2+b2ax1+by1+c,
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 a1x+b1y+c1=0 and a1x+b1y+c2=0 (same a1,b1), D=a12+b12∣c2−c1∣ — 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(x1,y1), the foot of the perpendicular on ax+by+c=0 is found from the parametric relation
ax−x1=by−y1=a2+b2−(ax1+by1+c),
and the image (reflection) of P in that line doubles the same offset:
ax−x1=by−y1=a2+b2−2(ax1+by1+c).
(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(x1,y1) lies on the origin side or non-origin side of ax+by+c=0 (c=0) according as ax1+by1+c has the same or opposite sign as c. …
Clear denominators to get 4x−3y−12=0, then use the perpendicular-distance formula from the origin.
The perpendicular distance of a point (x1,y1) from a line ax+by+c=0 is d=a2+b2∣ax1+by1+c∣. We first put the given intercept-form line into this general form.
Step 1. Clear the fractions. The line is 3x−4y=1. Multiply throughout by the LCM of 3 and 4, which is 12:
12(3x)−12(4y)=12⟹4x−3y=12⟹4x−3y−12=0
Step 2. Apply the perpendicular-distance formula from the origin (0,0). Here a=4,b=−3,c=−12: …