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Business Mathematics and Basic Statistics · Ch 5 — Coordinate Geometry (Two Dimensions)

Distance Between Two Points

2

Distance Between Two Points

Suppose two branches of a business, or two stops on a delivery route, sit at points P(x1,y1)P(x_1, y_1) and Q(x2,y2)Q(x_2, y_2) on the coordinate plane. How far apart are they, measured as a straight line — “as the crow flies”? This is answered by the distance formula, and it follows directly from the Pythagorean theorem already familiar from geometry.

Draw the segment PQPQ. Through PP draw a horizontal line and through QQ draw a vertical line, and let them meet at a point RR. Since PRPR is horizontal, its length is simply the difference of the x-coordinates, ∣x2−x1∣|x_2 - x_1|; since RQRQ is vertical, its length is the difference of the y-coordinates, ∣y2−y1∣|y_2 - y_1|. The angle at RR is a right angle, so triangle PRQPRQ is right-angled at RR, with PQPQ as its hypotenuse.

Figure 2 — Right triangle PRQ used to derive the distance formula, with P(2,3), Q(6,6), right-angle vertex R(6,3), and sides PR=4, RQ=3, hypotenuse PQ=5
Figure 2 — Right triangle PRQ used to derive the distance formula, with P(2,3), Q(6,6), right-angle vertex R(6,3), and sides PR=4, RQ=3, hypotenuse PQ=5

By the Pythagorean theorem,

PQ2=PR2+RQ2=(x2−x1)2+(y2−y1)2PQ^2 = PR^2 + RQ^2 = (x_2-x_1)^2 + (y_2-y_1)^2

so, taking the positive square root (a distance is never negative),

Note

Distance Formula

PQ=(x2−x1)2+(y2−y1)2PQ = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}

A useful special case follows by putting x1=y1=0x_1 = y_1 = 0: the distance of a point P(x,y)P(x, y) from the origin O(0,0)O(0, 0) is

OP=x2+y2OP = \sqrt{x^2 + y^2} …