# How do you write an equation (a) in slope-intercept form and (b) in standard form for the line passing through (1.7) and perpendicular to 3x+7y= 1?

Nov 25, 2016

The slope-intercept form is $y = \frac{7}{3} x + \frac{14}{3}$ and standard form is $- 7 x + 3 y = 14$.

#### Explanation:

1. Find the slope for $3 x + 7 y = 1$.
The slope of a line in $A x + B y = C$ form is always $- \frac{A}{B}$. (Rearranging $A x + B y = C$ into slope-intercept form gives $y = - \frac{A}{B} x + \frac{C}{B} .$) Thus, the slope of this line is $- \frac{3}{7}$.

2. Find the slope of a line perpendicular to this given line.
Lines that are perpendicular have slopes that are negative reciprocals of each other. (One line's $\text{rise"/"run}$ will be the other's (-"run")/"rise".) So the slope of a line perpendicular to this one will be $m = \frac{7}{3}$.

3. Plug in the known point $\left({x}_{1} , {y}_{1}\right)$ and $m$ into the slope-point equation $y - {y}_{1} = m \left(x - {x}_{1}\right)$.
Assuming the point given is $\left({x}_{1} , {y}_{1}\right) = \left(1 , 7\right)$, our slope-point equation is $y - 7 = \frac{7}{3} \left(x - 1\right)$.

4. Rearrange this equation into slope-intercept form.
Solving for $y$ yields:

$y = \frac{7}{3} x - \frac{7}{3} + 7$
$\implies y = \frac{7}{3} x + \frac{14}{3}$.

5. Rearrange (3) or (4) into standard form.
Starting from the equation in (3), we get

$3 y - 21 = 7 x - 7$
$\implies - 7 x + 3 y = 14$

(Alternatively, starting from (4), we get

$3 y = 7 x + 14$
$\implies - 7 x + 3 y = 14$

which is the same equation.)

A graph of $3 x + 7 y = 1$:
graph{3x+7y=1 [-12.74, 9.76, -2.16, 9.095]}
For every 3 it goes down, it goes 7 right.

A graph of $- 7 x + 3 y = 14$:
graph{-7x+3y=14 [-12.74, 9.76, -2.16, 9.095]}
For every 7 it goes up, it goes 3 right.
And as you can see, this second line does indeed pass through $\left(1 , 7\right)$.