How do you write an equation of a line with point (2,-3), slope 2/3?

Nov 3, 2017

See a solution process below:

Explanation:

We can use the point-slope formula for writing the equation for the line in the problem. The point-slope form of a linear equation is: $\left(y - \textcolor{b l u e}{{y}_{1}}\right) = \textcolor{red}{m} \left(x - \textcolor{b l u e}{{x}_{1}}\right)$

Where $\left(\textcolor{b l u e}{{x}_{1}} , \textcolor{b l u e}{{y}_{1}}\right)$ is a point on the line and $\textcolor{red}{m}$ is the slope.

Substituting the slope and values from the point in the problem gives:

$\left(y - \textcolor{b l u e}{- 3}\right) = \textcolor{red}{\frac{2}{3}} \left(x - \textcolor{b l u e}{2}\right)$

$\left(y + \textcolor{b l u e}{3}\right) = \textcolor{red}{\frac{2}{3}} \left(x - \textcolor{b l u e}{2}\right)$

We can solve this equation for $y$ to put the equation in slope-intercept format. The slope-intercept form of a linear equation is: $y = \textcolor{red}{m} x + \textcolor{b l u e}{b}$

Where $\textcolor{red}{m}$ is the slope and $\textcolor{b l u e}{b}$ is the y-intercept value.

$y + \textcolor{b l u e}{3} = \left(\textcolor{red}{\frac{2}{3}} \times x\right) - \left(\textcolor{red}{\frac{2}{3}} \times \textcolor{b l u e}{2}\right)$

$y + \textcolor{b l u e}{3} = \textcolor{red}{\frac{2}{3}} x - \frac{4}{3}$

$y + \textcolor{b l u e}{3} - 3 = \textcolor{red}{\frac{2}{3}} x - \frac{4}{3} - 3$

$y + 0 = \textcolor{red}{\frac{2}{3}} x - \frac{4}{3} - \left(\frac{3}{3} \times 3\right)$

$y = \textcolor{red}{\frac{2}{3}} x - \frac{4}{3} - \frac{9}{3}$

$y = \textcolor{red}{\frac{2}{3}} x - \textcolor{b l u e}{\frac{13}{3}}$