# Line L has equation 2x- 3y=5. Line M passes through the point (3, -10) and is parallel to line L. How do you determine the equation for line M?

Jun 15, 2017

See a solution process below:

#### Explanation:

Line L is in Standard Linear form. The standard form of a linear equation is: $\textcolor{red}{A} x + \textcolor{b l u e}{B} y = \textcolor{g r e e n}{C}$

Where, if at all possible, $\textcolor{red}{A}$, $\textcolor{b l u e}{B}$, and $\textcolor{g r e e n}{C}$are integers, and A is non-negative, and, A, B, and C have no common factors other than 1

$\textcolor{red}{2} x - \textcolor{b l u e}{3} y = \textcolor{g r e e n}{5}$

The slope of an equation in standard form is: $m = - \frac{\textcolor{red}{A}}{\textcolor{b l u e}{B}}$

Substituting the values from the equation into the slope formula gives:

$m = \frac{\textcolor{red}{- 2}}{\textcolor{b l u e}{- 3}} = \frac{2}{3}$

Because line M is parallel to line L, Line M will have the same slope.

We can now use the point-slope formula to write an equation for Line M. The point-slope formula states: $\left(y - \textcolor{red}{{y}_{1}}\right) = \textcolor{b l u e}{m} \left(x - \textcolor{red}{{x}_{1}}\right)$

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

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

$\left(y - \textcolor{red}{- 10}\right) = \textcolor{b l u e}{\frac{2}{3}} \left(x - \textcolor{red}{3}\right)$

$\left(y + \textcolor{red}{10}\right) = \textcolor{b l u e}{\frac{2}{3}} \left(x - \textcolor{red}{3}\right)$

If necessary for the answer we can transform this equation to the Standard Linear form as follows:

$y + \textcolor{red}{10} = \left(\textcolor{b l u e}{\frac{2}{3}} \times x\right) - \left(\textcolor{b l u e}{\frac{2}{3}} \times \textcolor{red}{3}\right)$

$y + \textcolor{red}{10} = \frac{2}{3} x - 2$

$\textcolor{b l u e}{- \frac{2}{3} x} + y + \textcolor{red}{10} - 10 = \textcolor{b l u e}{- \frac{2}{3} x} + \frac{2}{3} x - 2 - 10$

$- \frac{2}{3} x + y + 0 = 0 - 12$

$- \frac{2}{3} x + y = - 12$

$\textcolor{red}{- 3} \left(- \frac{2}{3} x + y\right) = \textcolor{red}{- 3} \times - 12$

$\left(\textcolor{red}{- 3} \times - \frac{2}{3} x\right) + \left(\textcolor{red}{- 3} \times y\right) = 36$

$\textcolor{red}{2} x - \textcolor{b l u e}{3} y = \textcolor{g r e e n}{36}$