# How do you write an equation of a line passing through (2, 3), perpendicular to y = (3/2)x − 4?

Apr 24, 2018

We find the line with negative reciprocal slope through the point, which is $y = - \frac{2}{3} x + \frac{13}{3}$. Another way to go is to swap the coefficients on $x$ and $y$ and negate one, and compute the constant from the point, giving $2 x + 3 y = 13$ which is the same line.

#### Explanation:

When we multiply perpendicular slopes together we get $- 1 ,$ meaning perpendicular slopes are the negative reciprocals of each other.

We can read off the slope of $y = \frac{3}{2} x - 4$ as the coefficient on $x$, which is $\frac{3}{2}$. The slope of the perpendiculars will be the negative reciprocal, $- \frac{2}{3.}$

We fit the constant to the line.

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

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

$b = \frac{13}{3}$

So the equation we seek is

$y = - \frac{2}{3} x + \frac{13}{3}$

I prefer not to deal with the slope if I can avoid it. I like first writing the line in standard form:

$- 3 x + 2 y = - 8$

Then the perpendiculars are gotten by swapping the coefficients on $x$ and $y$, and negating one. The constant is quickly determined by the point:

$2 x + 3 y = 2 \left(2\right) + 3 \left(3\right) = 13$

$2 x + 3 y = 13$