# Math Help - Equation of the Line

1. ## Equation of the Line

Show that an equation for a line with nonzero x-and y-intercepts can be written as (x/a) + (y/b) = 1 where a is the x-intercept and
b is the y-intercept. This is called the INTERCEPT FORM of the equation of a line.

2. Originally Posted by symmetry
Show that an equation for a line with nonzero x-and y-intercepts can be written as (x/a) + (y/b) = 1 where a is the x-intercept and
b is the y-intercept. This is called the INTERCEPT FORM of the equation of a line.
The x-intercept will be of the form (a, 0). Does this fit the equation?
$\frac{a}{a} + \frac{0}{b} = 1$ Check!

The y-intercept will be of the form (0, b). Does this fit the equation?
$\frac{0}{a} + \frac{b}{b} = 1$ Check!

If you have a need to show that this equation is indeed a line, then you can solve it for y:
$\frac{x}{a} + \frac{y}{b} = 1$

$\frac{y}{b} = -\frac{x}{a} + 1$

$y = - \left ( \frac{b}{a} \right ) x + b$

which is the slope-intercept form for a line.

-Dan

3. ## ok

So, I simply needed to set y = 0 to find the x-intercepts and set x = 0 to find the y-intercepts.

Thanks!

4. Originally Posted by symmetry
So, I simply needed to set y = 0 to find the x-intercepts and set x = 0 to find the y-intercepts.

Thanks!
Yup! What a lot of students tend to forget is that the y - intercept, for example, is a point. So when we say the y - intercept is "b" we are using a short-hand and being a bit sloppy: we really should be saying the y - intercept is the point (0, b). The same goes for the x - intercept.

-Dan

5. Hello, symmetry!

You may be expect to derive that formula . . .

Show that an equation for a line with nonzero x-and y-intercepts
can be written as: . $\frac{x}{a} + \frac{y}{b}\: = \:1$
where $a$ is the x-intercept and $b$ is the y-intercept.

We are given two points on the line: . $(a,0)$ and $(0,b)$.

The slope of the line is: . $m \:=\:\frac{b-0}{0-a} \:=\:-\frac{b}{a}$

The line through $(0,b)$ with slope $-\frac{b}{a}$ is:
. . . . $y - b \:=\:-\frac{b}{a}(x - 0)\quad\Rightarrow\quad y \:=\:-\frac{b}{a}x + b\quad\Rightarrow\quad \frac{bx}{a} + y \:=\:b$

Divide through by $b\!:\;\;\boxed{\frac{x}{a} + \frac{y}{b}\:=\:1}$