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

Dec 30, 2016

In slope-point form, the equation of line M is $y - 10 = - \frac{3}{2} \left(x - 2\right)$.
In slope-intercept form, it is $y = - \frac{3}{2} x + 13$.

#### Explanation:

In order to find the slope of line M, we must first deduce the slope of line L.

The equation for line L is $2 x - 3 y = 5$. This is in standard form, which does not directly tell us the slope of L. We can rearrange this equation, however, into slope-intercept form by solving for $y$:

$2 x - 3 y = 5$

$\textcolor{w h i t e}{2 x} - 3 y = 5 - 2 x \text{ }$(subtract $2 x$ from both sides)

$\textcolor{w h i t e}{2 x - 3} y = \frac{5 - 2 x}{- 3} \text{ }$(divide both sides by $- 3$)

$\textcolor{w h i t e}{2 x - 3} y = \frac{2}{3} x - \frac{5}{3} \text{ }$(rearrange into two terms)

This is now in slope-intercept form $y = m x + b$, where $m$ is the slope and $b$ is the $y$-intercept. So, the slope of line L is $\frac{2}{3}$.

(Incidentally, since the slope of $2 x - 3 y = 5$ was found to be $\frac{2}{3}$, we can show that the slope of any line $A x + B y = C$ will be $- \frac{A}{B}$. This may be useful to remember.)

Okay. Line M is said to be perpendicular to line L—that is, lines L and M create right angles where they cross.

The slopes of two perpendicular lines will be negative reciprocals of each other. What does this mean? It means that if the slope of a line is $\frac{a}{b}$, then the slope of a perpendicular line will be $- \frac{b}{a}$.

Since the slope of line L is $\frac{2}{3}$, the slope of line M will be $- \frac{3}{2}$.

Alright—now we know the slope of line M is $- \frac{3}{2}$, and we know a point that it passes through: $\left(2 , 10\right)$. We now simply pick an equation for a line that allows us to plug in this data. I will choose to insert the data into the slope-point equation for a line:

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

$y - 10 = - \frac{3}{2} \left(x - 2\right)$

Choosing slope-point form allows us to simply stop here. (You could choose to use $y = m x + b$, where $\left(x , y\right) = \left(2 , 10\right)$ and $m = - \frac{3}{2}$, then solve for $b$, and finally use this $b$ along with $m$ in slope-intercept form again:

$y = \text{ "mx" } + b$

$10 = - \frac{3}{2} \left(2\right) + b$

$10 = \text{ "-3" } + b$

$13 = b$

$\therefore y = m x + b$
$\implies y = - \frac{3}{2} x + 13$

Same line, different form.)