# Parameterize the line through P = (8,-3) and Q = (10,0) so that the points P and correspond to...

## Question:

Parameterize the line through P = (8,-3) and Q = (10,0) so that the points P and correspond to the parameter values t = 5 and 8.

## Parametric Equations of a Line

A line in two and three dimensional space can be represented by a parametric function. In two dimensional space the function has the form {eq}\textbf{r}:\left[a,b\right]\rightarrow\mathbb{R}^2 {/eq} defined as {eq}\textbf{r}(t)=(1-t)\textbf{A}+t\textbf{B} {/eq} where {eq}\textbf{A}=(a,b) {/eq} and {eq}\textbf{B}=(c,d) {/eq} are points on the line. The definition in three dimensional space is a direct extension. In the question the lines and points are in two dimensional space. In the question points {eq}\textbf{A} {/eq} and {eq}\textbf{B} {/eq} must be found so that two given points on the line correspond to two given values of the parameter {eq}t {/eq}. Finding the points results in solving two linear systems of equations each having two unknowns.

## Answer and Explanation:

The question is restated with slightly different notation. Parametrize the line through {eq}\textbf{P}=(8,-3) {/eq} and {eq}\textbf{Q}=(10,0) {/eq} so that the points {eq}\textbf{P} {/eq} and {eq}\textbf{Q} {/eq} correspond to the parameter values {eq}t=5 {/eq} and {eq}t=8 {/eq}, respectively.

The line can be expressed as the following parametric function {eq}\textbf{r}:\left[0,1\right]\rightarrow\mathbb{R}^2 {/eq} defined as

{eq}\begin{eqnarray*} \textbf{r}(t) &=& (1-t)\textbf{A}+t\textbf{B} \\ \textbf{r}(t) &=& (1-t)(a,b)+t(c,d) \\ &=& (x,y) \end{eqnarray*} {/eq}

where {eq}\textbf{A}=(a,b) {/eq} and {eq}\textbf{B}=(c,d) {/eq} are points on the line which satisfy

{eq}\begin{eqnarray*} \textbf{P} &=& (8,-3) \\ \textbf{r}(5) &=& (1-5)\textbf{A}+5\textbf{B} \\ \textbf{r}(5) &=& (1-5)(a,b)+5(c,d) \\ &=& -4(a,b)+5(c,d) \\ &=& (-4a,-4b)+(5c,5d) \\ &=& (-4a+5c,-4b+5d) \\ \textbf{Q} &=& (10,0) \\ \textbf{r}(8) &=& (1-8)\textbf{A}+8\textbf{B} \\\ \textbf{r}(8) &=& (1-8)(a,b)+8(c,d) \\ &=& -7(a,b)+8(c,d) \\ &=& (-7a,-7b)+(8c,8d) \\ &=& (-7a+8c,-7b+8d). \end{eqnarray*} {/eq}

The above equations result in the linear system of equations

{eq}\begin{eqnarray*} 8 &=& -4a+5c \\ 10 &=& -7a+8c \end{eqnarray*} {/eq}

in the unknowns {eq}a {/eq} and {eq}c {/eq} and the linear system of equations

{eq}\begin{eqnarray*} -3 &=& -4b+5d \\ 0 &=& -7b+8d \end{eqnarray*} {/eq}

in the unknowns {eq}b {/eq} and {eq}d {/eq}.

Consider the first linear system in the unknowns {eq}a {/eq} and {eq}c {/eq}. Multiply the first equation in the system by 7 and the second equation by {eq}-4 {/eq} to get

{eq}\begin{eqnarray*} 56 &=& -28a+35c \\ -40 &=& 28a-32c. \end{eqnarray*} {/eq}

Add the above equations to eliminate {eq}a {/eq} and solve for {eq}c {/eq} to get {eq}c=\frac{16}{3} {/eq}. Plug {eq}c=\frac{16}{3} {/eq} into the first equation in the original system and solve for {eq}a {/eq} to get

{eq}\begin{eqnarray*} 8 &=& -4a+5\left(\frac{16}{3}\right) \\ 2 &=& -a+5\left(\frac{4}{3}\right) \\ 6 &=& -3a+20 \\ -14 &=& -3a \\ \frac{14}{3} &=& a. \end{eqnarray*} {/eq}

Consider the second linear system in the unknowns {eq}b {/eq} and {eq}d {/eq}. Multiply the first equation in the system by {eq}-7 {/eq} and the second equation by 4 to get

{eq}\begin{eqnarray*} 21 &=& 28b-35d \\ 0 &=& -28b+32d. \end{eqnarray*} {/eq}

Add the above equations to eliminate {eq}b {/eq} and solve for {eq}d {/eq} to get {eq}d=-7 {/eq}. Plug {eq}d=-7 {/eq} into the second equation in the original system and solve for {eq}b {/eq} to get

{eq}\begin{eqnarray*} 0 &=& -7b+8(-7) \\ 0 &=& b+8 \\ -8 &=& b. \end{eqnarray*} {/eq}

The desired points are {eq}\textbf{A}=\left(\frac{14}{3},-8\right) {/eq} and {eq}\textbf{B}=\left(\frac{16}{3},-7\right) {/eq}. The desired parametrization is therefore

{eq}\begin{eqnarray*} \textbf{r}(t) &=& (1-t)\left(\frac{14}{3},-8\right)+t\left(\frac{16}{3},-7\right). \end{eqnarray*} {/eq}

**Check**: As a check note that

{eq}\begin{eqnarray*} \textbf{r}(5) &=& (1-5)\left(\frac{14}{3},-8\right)+5\left(\frac{16}{3},-7\right) \\ &=& -4\left(\frac{14}{3},-8\right)+5\left(\frac{16}{3},-7\right) \\ &=& \left(-\frac{56}{3},32\right)+\left(\frac{80}{3},-35\right) \\ &=& \left(\frac{24}{3},-3\right) \\ &=& (8,-3) \\ &=& \textbf{P} \\ \textbf{r}(8) &=& (1-8)\left(\frac{14}{3},-8\right)+8\left(\frac{16}{3},-7\right) \\ &=& -7\left(\frac{14}{3},-8\right)+8\left(\frac{16}{3},-7\right) \\ &=& \left(-\frac{98}{3},56\right)+\left(\frac{128}{3},-56\right) \\ &=& \left(\frac{30}{3},0\right) \\ &=& (10,0) \\ &=& \textbf{Q}. \end{eqnarray*} {/eq}

#### Learn more about this topic:

from Precalculus: High School

Chapter 24 / Lesson 3