Parametric Equations Math Example 3

Follow the full solution, then compare it with the other examples linked below.

Example 3

medium
Find parametric equations for the line through (1,4)(1, 4) and (5,−2)(5, -2).

Solution

  1. 1
    The direction vector is (5−1,−2−4)=(4,−6)(5-1, -2-4) = (4, -6). Using point (1,4)(1, 4): x=1+4tx = 1 + 4t, y=4−6ty = 4 - 6t.
  2. 2
    Verify: at t=0t = 0, (x,y)=(1,4)(x,y) = (1,4) ✓; at t=1t = 1, (x,y)=(5,−2)(x,y) = (5,-2) ✓.

Answer

x=1+4t,y=4−6tx = 1 + 4t, \quad y = 4 - 6t
Parametric equations for a line use a point and a direction vector: (x,y)=(x0,y0)+t(a,b)(x, y) = (x_0, y_0) + t(a, b). The parameter t=0t = 0 gives the starting point, and varying tt traces the line. Different choices of point and direction give equivalent parameterizations.

About Parametric Equations

A way of defining a curve by expressing both xx and yy as separate functions of a third variable (parameter), typically tt: x=f(t)x = f(t), y=g(t)y = g(t).

Learn more about Parametric Equations →

More Parametric Equations Examples

Example 1 easy

Eliminate the parameter from [formula] and [formula] to find the rectangular equation.

Example 2 medium

Eliminate the parameter from [formula] and [formula].

Example 4 hard

Eliminate the parameter from [formula] and [formula] and describe the curve.

← All Parametric Equations Examples Parametric Equations Concept