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

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.

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 $y=x$.

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" $\mathrm{y}=3-\mathrm{x}$.

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$y=\frac{6}{5}+\frac{x}{5}$

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 $\mathrm{y}=\frac{15}{4}+\frac{9\mathrm{x}}{8}$.