Lines in the Plane

Increments and Distance Between Points In The Plane

If we move from a point P = (x_1,y_1) to a point Q = (x_2,y_2) in the plane, then we will have two increments or changes to consider. The increment in the x or horizontal direction is x_2 – x_1 which is denoted by ∆x = x_2 – x_1. The increment in the y or vertical direction is ∆y = y_2 – y_1. These increments are shown in Fig. 5 . ∆x does not represent ∆ times x, it represents the difference in the x coordinates: ∆x = x_2 – x_1.


The distance between the points P = (x_1,y_1) and Q = (x_2,y_2) is simply an application of the Pythagorean formula for right triangles, and

dist(P,Q) = \sqrt{(∆x)^2 + (∆y)^2} = \sqrt{ (x_2–x_1)^2 + (y_2–y_1)^2}

The midpoint M of the line segment joining P and Q is M = (\dfrac{x_1 + x_2}{2} , \dfrac{y_1 + y_2}{2})

Example 2: Find an equation describing the points P = (x,y) which are equidistant from Q = (2,3) and R = (5,–1). (Fig. 6) 


Solution: The points P=(x,y) must satisfy dist(P,Q) = dist(P,R) so

\sqrt{(x–2)^2+(y–3)^2} =\sqrt {(x–5)^2+(y–(–1))^2}

By squaring each side we get (x–2)^2+(y–3)^2 = (x–5)^2+(y+1)^2

Then x^2 – 4x + 4 + y^2 – 6y + 9 = x^2 – 10x + 25 + y^2 + 2y + 1

so –4x – 6y + 13 = –10x + 2y + 26 and y = .75x – 1.625, a straight line. Every point on the line y = .75x – 1.625 is equally distant from Q and R.

Practice 2: Find an equation describing all points P = (x,y) equidistant from Q = (1,–4) and R = (0,–3).

A circle with radius r and center at the point C = (a,b) consists of all points P = (x,y) which are at a distance of r from the center C: the points P which satisfy dist(P,C) = r.

Example 3: Find the equation of a circle with radius r = 4 and center C = (5,–3). (Fig. 7)


Solution: A circle is the set of points P=(x,y) which are at a fixed distance r from the center point C, so this circle will be the set of points P=(x,y) which are at a distance of 4 units from the point C = (5,–3). P will be on this circle if dist(P,C) = 4.
Using the distance formula and simplifying,

\sqrt{(x–5)^2 + (y+3)^2} = 4 so (x–5)^2 + (y+3)^2=16 or

x^2 – 10x + 25 + y^2 + 6y + 9 = 16.

Practice 3: Find the equation of a circle with radius r = 5 and center C = (–2,6).