Question

The equation for a straight line (deterministic model) is

$\mathit{y}\mathbf{=}{\mathit{\beta }}_{\mathbf{0}}\mathbf{+}{\mathit{\beta }}_{\mathbf{1}}\mathit{x}$

If the line passes through the point (-2, 4), then x = -2, y = 4 must satisfy the equation; that is,

$\mathbf{4}\mathbf{=}{\mathit{\beta }}_{\mathbf{0}}\mathbf{+}{\mathit{\beta }}_{\mathbf{1}}\mathbf{(}\mathbf{-}\mathbf{2}\mathbf{)}$

Similarly, if the line passes through the point (4, 6), then x = 4, y = 6 must satisfy the equation; that is,

$\mathbf{6}\mathbf{=}{\mathit{\beta }}_{\mathbf{0}}\mathbf{+}{\mathit{\beta }}_{\mathbf{1}}\mathbf{\left(}\mathbf{4}\mathbf{\right)}$

Use these two equations to solve for and ; then find the equation of the line that passes through the points (-2, 4) and (4, 6).

Short answer

${\mathit{\beta }}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{14}}{\mathbf{3}}$, ${\mathit{\beta }}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{-}\mathbf{1}}{\mathbf{3}}$

3y = 14 + x

Step-by-step solution

Introduction

In geometry, a line is essentially a zero-width object that extends on both sides. A straight line is a line that does not have any curves. A straight line is one that has no curves and extends to both sides to infinity.

Finding  β0 and β1

Equation of a straight line:

$\mathit{y}\mathbf{=}{\mathit{\beta }}_{\mathbf{0}}\mathbf{+}{\mathit{\beta }}_{\mathbf{1}}\mathit{x}$

If a line passes through, (2, 4),

$\mathbf{4}\mathbf{=}{\mathit{\beta }}_{\mathbf{0}}\mathbf{+}{\mathit{\beta }}_{\mathbf{1}}\mathbf{(}\mathbf{-}\mathbf{2}\mathbf{)}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

If a line passes through (4, 6),

$\mathbf{6}\mathbf{=}{\mathit{\beta }}_{\mathbf{0}}\mathbf{+}{\mathit{\beta }}_{\mathbf{1}}\mathbf{\left(}\mathbf{4}\mathbf{\right)}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

Solving equations (1) & (2) simultaneously, we get

$$\begin{gathered} {\mathit{\beta }}_{\mathbf{0}}\mathbf{+}{\mathit{\beta }}_{\mathbf{1}}\mathbf{(}\mathbf{-}\mathbf{2}\mathbf{)}\mathbf{-}\mathbf{4}\mathbf{=}{\mathit{\beta }}_{\mathbf{0}}\mathbf{+}{\mathit{\beta }}_{\mathbf{1}}\mathbf{\left(}\mathbf{4}\mathbf{\right)}\mathbf{-}\mathbf{6} \\ \mathbf{6}{\mathit{\beta }}_{\mathbf{1}}\mathbf{=}\mathbf{-}\mathbf{2} \\ {\mathit{\beta }}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{-}\mathbf{2}}{\mathbf{6}} \\ {\mathit{\beta }}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{-}\mathbf{1}}{\mathbf{3}} \end{gathered}$$

Therefore, we get

$$\begin{gathered} \mathbf{4}\mathbf{=}{\mathit{\beta }}_{\mathbf{0}}\mathbf{+}\frac{\mathbf{-}\mathbf{1}}{\mathbf{3}}\mathbf{(}\mathbf{-}\mathbf{2}\mathbf{)} \\ \mathbf{4}\mathbf{+}\frac{\mathbf{2}}{\mathbf{3}}\mathbf{=}{\mathit{\beta }}_{\mathbf{0}} \\ {\mathit{\beta }}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{12}\mathbf{+}\mathbf{2}}{\mathbf{3}} \\ {\mathit{\beta }}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{14}}{\mathbf{3}} \end{gathered}$$

Finding the equation that passes through the points (-2, 4) and (4, 6) 

$$\begin{gathered} \mathit{y}\mathbf{=}{\mathit{\beta }}_{\mathbf{0}}\mathbf{+}{\mathit{\beta }}_{\mathbf{1}}\mathit{x} \\ \mathit{y}\mathbf{=}\frac{\mathbf{14}}{\mathbf{3}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{3}}\mathit{x} \\ \mathit{y}\mathbf{=}\frac{\mathbf{14}\mathbf{+}\mathbf{x}}{\mathbf{3}} \\ \mathbf{3}\mathit{y}\mathbf{=}\mathbf{14}\mathbf{+}\mathit{x} \end{gathered}$$

Therefore, the required equation is 3y = 14 + x.