Question

Refer to Exercise 11.3. Find the equations of the lines that pass through the points listed in Exercise 11.1.

Short answer

1. y = x
2. y = 3 – x
3. $\mathrm{y}=\frac{6}{5}+\frac{\mathrm{x}}{5}$
4. $\mathrm{y}=\frac{15}{4}+\frac{9\mathrm{x}}{8}$
   

Step-by-step solution

Introduction

Any x-y coordinate plane can be used to graph equations involving one or two variables. The following guidelines are valid in general: The coordinates of a point on the graph of an equation make the equation true, if a point's coordinates result in a true statement for an equation, the point is on the equation's graph.

Determine the equation of the line passing through the point (1, 1) and (5, 5).

Equation of straight line:

$\mathrm{y}={\mathrm{\beta }}_0+{\mathrm{\beta }}_1\mathrm{x}.............\left(1\right)$

If line passing through, (1, 1).

$1={\mathrm{\beta }}_0+{\mathrm{\beta }}_1\left(1\right)..............\left(2\right)$

If line passing through, (5, 5).

$5={\mathrm{\beta }}_0+{\mathrm{\beta }}_1\left(5\right)..............\left(3\right)$

Solve equation (2) & (3) simultaneously,

$$\begin{gathered} {\mathrm{\beta }}_0+{\mathrm{\beta }}_1-1={\mathrm{\beta }}_0+{\mathrm{\beta }}_1\left(5\right)-5 \\ {\mathrm{\beta }}_1-1=5{\mathrm{\beta }}_1-5 \\ 4{\mathrm{\beta }}_1=4 \\ {\mathrm{\beta }}_1=1 \end{gathered}$$

Therefore, put ${\mathrm{\beta }}_1=1$ in equation (2)

$$\begin{gathered} 1={\mathrm{\beta }}_0+1\left(1\right) \\ 1={\mathrm{\beta }}_0+1 \\ {\mathrm{\beta }}_0=0 \end{gathered}$$

Now, put ${\mathrm{\beta }}_1=1$and ${\mathrm{\beta }}_1=0$ in equation (3)

role="math" localid="1668667810631" $$\begin{gathered} \mathrm{y}={\mathrm{\beta }}_0+{\mathrm{\beta }}_1\mathrm{x} \\ \mathrm{y}=0+1\mathrm{x} \\ \mathrm{y}=\mathrm{x} \end{gathered}$$

Therefore, the required equation is $y=x$.

Determine the equation of the line passing through the point (0, 3) and (3, 0).

Equation of straight line:

$\mathrm{y}={\mathrm{\beta }}_0+{\mathrm{\beta }}_1\mathrm{x}..............\left(1\right)$

If line passing through, (0, 3).

$$\begin{gathered} 3={\mathrm{\beta }}_0+{\mathrm{\beta }}_1\left(0\right) \\ {\mathrm{\beta }}_0=3 \end{gathered}$$

If line passing through, (3, 0).

$0={\mathrm{\beta }}_0+{\mathrm{\beta }}_1\left(3\right)............\left(2\right)$

Therefore, put ${\mathrm{\beta }}_0=3$ in equation (2)

$$\begin{gathered} 0=3+3{\mathrm{\beta }}_1 \\ 3{\mathrm{\beta }}_1=-3 \\ {\mathrm{\beta }}_1=-1 \end{gathered}$$

Now, put ${\mathrm{\beta }}_1=-1$and ${\mathrm{\beta }}_0=3$ in equation (3)

$$\begin{gathered} \mathrm{y}={\mathrm{\beta }}_0+{\mathrm{\beta }}_1\mathrm{x} \\ \mathrm{y}=3+\left(-1\right)\mathrm{x} \\ \mathrm{y}=3-\mathrm{x} \end{gathered}$$

Therefore, the required equation isrole="math" localid="1668668119632" $\mathrm{y}=3-\mathrm{x}$.

Determine the equation of the line passing through the point (-1, 1) and (4, 2).

Equation of straight line:

$\mathrm{y}={\mathrm{\beta }}_0+{\mathrm{\beta }}_1\mathrm{x}..............\left(1\right)$

If line passing through, (-1, 1).

$1={\mathrm{\beta }}_0+{\mathrm{\beta }}_1\left(-1\right).........\left(2\right)$

If line passing through, (4, 2).

$2={\mathrm{\beta }}_0+{\mathrm{\beta }}_1\left(4\right).........\left(3\right)$

Solve equation (2) & (3) simultaneously,

$$\begin{gathered} {\mathrm{\beta }}_0+{\mathrm{\beta }}_1(-1)-1={\mathrm{\beta }}_0+{\mathrm{\beta }}_1\left(4\right)-2 \\ -{\mathrm{\beta }}_1-1=4{\mathrm{\beta }}_1-2 \\ 4{\mathrm{\beta }}_1+{\mathrm{\beta }}_1=2-1 \\ 5{\mathrm{\beta }}_1=1 \\ {\mathrm{\beta }}_1=\frac{1}{5} \end{gathered}$$

Therefore, put ${\mathrm{\beta }}_1=\frac{1}{5}$ in equation (2)

$$\begin{gathered} 1={\mathrm{\beta }}_0+\frac{1}{5}(-1) \\ 1={\mathrm{\beta }}_0-\frac{1}{5} \\ {\mathrm{\beta }}_0=1+\frac{1}{5} \\ {\mathrm{\beta }}_0=\frac{5+1}{5} \\ {\mathrm{\beta }}_0=\frac{6}{5} \end{gathered}$$

Now, put ${\mathrm{\beta }}_0=\frac{1}{5}$ and ${\mathrm{\beta }}_0=\frac{6}{5}$ in equation (3)

$$\begin{gathered} \mathrm{y}={\mathrm{\beta }}_0+{\mathrm{\beta }}_1\mathrm{x} \\ \mathrm{y}=\frac{6}{5}+\frac{1}{5}\mathrm{x} \\ \mathrm{y}=\frac{6}{5}+\frac{\mathrm{x}}{5} \end{gathered}$$

Therefore, the required equation is$y=\frac{6}{5}+\frac{x}{5}$

Determine the equation of the line passing through the point (-6, -3) and (2, 6).

Equation of straight line:

$\mathrm{y}={\mathrm{\beta }}_0+{\mathrm{\beta }}_1\mathrm{x}..............\left(1\right)$

If line passing through, (-6, -3).

$-3={\mathrm{\beta }}_0+{\mathrm{\beta }}_1(-6).........\left(2\right)$

If line passing through, (2, 6).

$6={\mathrm{\beta }}_0+{\mathrm{\beta }}_1\left(2\right).........\left(3\right)$

Solve equation (2) & (3) simultaneously,

$$\begin{gathered} {\mathrm{\beta }}_0+\left(-6\right){\mathrm{\beta }}_1+3={\mathrm{\beta }}_0+{\mathrm{\beta }}_1\left(2\right)-6 \\ -6{\mathrm{\beta }}_1+3=2{\mathrm{\beta }}_1-6 \\ 6{\mathrm{\beta }}_1+2{\mathrm{\beta }}_1=6+3 \\ 8{\mathrm{\beta }}_1=9 \\ {\mathrm{\beta }}_1=\frac{9}{8} \end{gathered}$$

Therefore, put ${\mathrm{\beta }}_1=\frac{9}{8}$ in equation (2)

$$\begin{gathered} -3={\mathrm{\beta }}_0+\frac{9}{8}\left(-6\right) \\ -3={\mathrm{\beta }}_0-\frac{54}{8} \\ {\mathrm{\beta }}_0=\frac{54}{8}-3 \\ {\mathrm{\beta }}_0=\frac{54-24}{8} \\ {\mathrm{\beta }}_0=\frac{30}{8} \\ {\mathrm{\beta }}_0=\frac{15}{4} \end{gathered}$$

Now, put ${\mathrm{\beta }}_1=\frac{9}{8}$and ${\mathrm{\beta }}_0=\frac{15}{4}$ in equation (3)

$$\begin{gathered} \mathrm{y}={\mathrm{\beta }}_0+{\mathrm{\beta }}_1\mathrm{x} \\ \mathrm{y}=\frac{15}{4}+\frac{9}{8}\mathrm{x} \\ \mathrm{y}=\frac{15}{4}+\frac{9\mathrm{x}}{8} \end{gathered}$$

Therefore, the required equation is $\mathrm{y}=\frac{15}{4}+\frac{9\mathrm{x}}{8}$.