The line looks like this: Since we like going from left to right, put *t* = 0 at the point (2, 3). Since *t* = 1 is a nice number as well, put* t* = 1 at the point (7, 9). We'll end with a parametrization that takes one time step to travel from one point to the other. We're going to look at *x* and *y* separately. We'll end with a linear equation expressing *x* in terms of *t* and a linear equation expressing *y* in terms of *t*. First deal with *x* and forget *y* ever existed. Write an equation expressing *x* as a line in terms of *t*. When* t* = 0 we have *x* = 2, and when *t* = 1 we have* x* = 7. The intercept of this line is 2 and its slope is The line expressing *x* in terms of *t* is
*x*(*t*) = 2 + 5*t*. Now forget about *x*. Write an equation expressing *y* as a line in terms of *t*. When *t* = 0 we have *y* = 3 and when *t* = 1 we have *y* = 9. The intercept is 3 and the slope is The line expressing *y* in terms of *t* is* y*(*t*) = 3 + 6*t*. The final parametrization is *x*(*t*) = 2 + 5*t*
*y*(*t*) = 3 + 6*t*
To check that this parametrization is correct we need to make sure that its graph goes through the points (2, 3) and (7 ,9). When *t* = 0 we have *x *= 2 + 5(0) = 2
*y* = 3 + 6(0) = 3
so the graph goes through (2,3). When *t* = 1 we have *x* = 2 + 5(1) = 7
*y* = 3 + 6(1) = 9
so the graph goes through the point (7, 9). That means we found a correct parameterization, and we've made a hot dog with a bun and relish. |