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.
Concurrency of lines. Three (or more) lines are concurrent if they all pass through one common point. The family of lines through the intersection of L1≡a1x+b1y+c1=0 and L2≡a2x+b2y+c2=0 is L1+λL2=0 for a parameter λ — this represents every line through their intersection point except L2=0 itself, and lets a line meeting one further condition (through a third point, parallel/perpendicular to a given line) be found by solving for λwithout first computing the intersection point. Three given lines are concurrent exactly when one of them can be written as L1+λL2=0 for some λ (equivalently, their 3×3 coefficient determinant vanishes).
Tip
A 'shortest cable/path connecting two points via a line/wall' problem (a substation on a road connecting two villages, an ant's shortest path around a cylinder or box) is solved by reflecting one of the two fixed points across the line/surface and joining the reflected point to the other by a straight segment — the straight segment's length is the shortest path, and the point where it crosses the line/surface is the optimal 'via' point. For a curved or faceted surface (a cylinder, the four walls of a box), first unroll/unfold the surface into a flat strip so the geometry becomes an ordinary straight-line-in-a-plane problem.
Distance from (x1,y1) to ax+by+c=0 is D=a2+b2ax1+by1+c; here a2+b2=16+9=5.
(i) ∣4(−2)+3(4)+4∣/5=8/5.
(ii) ∣4(7)+3(−3)+4∣/5=23/5.
✓Final answer
(i) 58 units (ii) 523 units
Distance from (x1,y1) to ax+by+c=0 is D=a2+b2ax1+by1+c; here a2+b2=16+9=5.
(i) ∣4(−2)+3(4)+4∣/5=8/5.
(ii) ∣4(7)+3(−3)+4∣/5=23/5.
Both parts use the standard perpendicular-distance formula from a point to a line, applied to the same line 4x+3y+4=0 but two different points.
Step 1. Set up the distance formula. For the line 4x+3y+4=0, a=4,b=3,c=4, so
D=42+324x1+3y1+4=54x1+3y1+4.
Step 2. Part (i): point (−2,4).
D=54(−2)+3(4)+4=5−8+12+4=58=58.
Step 3. Part (ii): point (7,−3).
D=54(7)+3(−3)+4=528−9+4=523=523.
✓Final answer
(i) 58 units (ii) 523 units
Forgetting the absolute value, which can leave a negative 'distance'.
Using a2+b2 from the wrong line (e.g. swapping a and b).