# How do you write an equation (a) in slope-intercept form and (b) in standard form for the line passing through (3,7) and perpendicular to 4x+5y=1?

Apr 21, 2018

Slope-intercept : $y = \frac{5}{4} x + \frac{13}{4}$
Standard : $- 5 x + 4 y = 13$

#### Explanation:

To write a linear equation in any form, you need at least two of the following three pieces of information: a point, a different point, and/or a slope. We are already given a single point, $\left(3 , 7\right)$.

Note that if we have a line $y = m x + b$ in slope-intercept form, its slope is given by $m$. The line perpendicular to this line will have a slope of $- \frac{1}{m}$. So let's put our provided equation in slope-intercept form.

$4 x + 5 y = 1$
$5 y = 1 - 4 x$
$y = \frac{1}{5} - \frac{4}{5} x = - \frac{4}{5} x + \frac{1}{5}$

The slope of the given line is then $- \frac{4}{5}$. The slope of the line perpendicular to this must be $\frac{5}{4}$. Now we have both a slope and a point.

To find our equation in slope-intercept form, we take the base form $y = m x + b$ and plug in our slope of $m = \frac{5}{4}$ and our point of $\left(x , y\right) = \left(3 , 7\right)$ to find $b$.

$y = m x + b$
$y = \frac{5}{4} x + b$
$7 = \frac{5}{4} \left(3\right) + b$
$7 = \frac{15}{4} + b$
$7 - \frac{15}{4} = b$
$\frac{28}{4} - \frac{15}{4} = b$
$\frac{13}{4} = b$.

Having found $b$, we now know our equation in slope-intercept form. That is, $y = \frac{5}{4} x + \frac{13}{4}$.

The standard form of a linear equation is $a x + b y = c$, where $a , b , c$ are constants. It is easy to move from slope-intercept form into standard form.

$y = \frac{5}{4} x + \frac{13}{4}$
$4 y = 5 x + 13$
$- 5 x + 4 y = 13$

Thus, our answer in standard form is $- 5 x + 4 y = 13$.