# Find a parametrization of the line segment starting at the point (x,y)=(-2,-6) and ending at the...

## Question:

Find a parametrization of the line segment starting at the point {eq}(x,y)=(-2,-6){/eq} and ending at the point {eq}(x,y)=(3,4){/eq}

using affine functions {eq}x(t){/eq} and {eq}y(t){/eq} for {eq}t \epsilon [0,1]{/eq}.

Note: {eq}f{/eq} is an affine function of {eq}t{/eq} if it can be written as {eq}f(t)=at+b{/eq}, for some constants {eq}a{/eq} and {eq}b{/eq}.

{eq}x={/eq}

for {eq}t \epsilon [0,1]{/eq}.

{eq}y={/eq}

for {eq}t \epsilon [0,1].{/eq}

## Affine Functions of a Variable:

If {eq}f(t) {/eq} is a function of the variable {eq}t {/eq}, then we say that {eq}f(t) {/eq} is affine if we can write {eq}f(t)=at+b {/eq} for some constants {eq}a {/eq} and {eq}b {/eq}.

If {eq}f {/eq} is affine, then:

• {eq}b {/eq} is the value of {eq}f {/eq} at {eq}t=0 {/eq}: that is, {eq}b=f(0) {/eq}
• {eq}a {/eq} is the rate of change of {eq}f {/eq} on any interval: that is, {eq}a=\dfrac{f(t_1)-f(t_0)}{t_1-t_0} {/eq} whenever {eq}t_0\ne t_1 {/eq}. In particular, this means that {eq}a=f(1)-f(0) {/eq}.

We'll look for a parameterization of this line segment of the form {eq}r(t)=(x(t),y(t))=(at+b,ct+d) {/eq}. Since we want {eq}r(0)=(-2,-6) {/eq} and {eq}r(1)=(3,4) {/eq}, we must have

{eq}\begin{align*} x(0)&=-2\\ x(1)&=3\\ y(0)&=-6\\ y(1)&=4 \, . \end{align*} {/eq}

Since {eq}x(0)=-2 {/eq} and {eq}x(t)=at+b {/eq}, we must have {eq}b=-2 {/eq}. Similarly, the rate of change of {eq}x(t) {/eq} is given by

{eq}\begin{align*} x(1)-x(0)&=3-(-2)\\ &=5 \end{align*} {/eq}

and so {eq}a=5 {/eq}. That is, {eq}x(t)=5t-2 {/eq}.

Since {eq}y(0)=-6 {/eq} and {eq}y(t)=ct+d {/eq}, we must have {eq}d=-6 {/eq}. Similarly, the rate of change of {eq}y(t) {/eq} is given by

{eq}\begin{align*} y(1)-y(0)&=4-(-6)\\ &=10 \end{align*} {/eq}

and so {eq}c=10 {/eq}. That is, {eq}y(t)=10t-6 {/eq}.

Putting all this together, the line segment has the parameterization {eq}\boxed{r(t)=(5t-2,10t-6)}\, {/eq}.