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?
$\displaystyle \frac{a}{a} + \frac{0}{b} = 1$ Check!
The y-intercept will be of the form (0, b). Does this fit the equation?
$\displaystyle \frac{0}{a} + \frac{b}{b} = 1$ Check!
If you have a need to show that this equation is indeed a line, then you can solve it for y:
$\displaystyle \frac{x}{a} + \frac{y}{b} = 1$
$\displaystyle \frac{y}{b} = -\frac{x}{a} + 1$
$\displaystyle y = - \left ( \frac{b}{a} \right ) x + b$
which is the slope-intercept form for a line.
-Dan
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
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: .$\displaystyle \frac{x}{a} + \frac{y}{b}\: = \:1$
where $\displaystyle a$ is the x-intercept and $\displaystyle b$ is the y-intercept.
We are given two points on the line: .$\displaystyle (a,0)$ and $\displaystyle (0,b)$.
The slope of the line is: .$\displaystyle m \:=\:\frac{b-0}{0-a} \:=\:-\frac{b}{a}$
The line through $\displaystyle (0,b)$ with slope $\displaystyle -\frac{b}{a}$ is:
. . . . $\displaystyle y - b \:=\:-\frac{b}{a}(x - 0)\quad\Rightarrow\quad y \:=\:-\frac{b}{a}x + b\quad\Rightarrow\quad \frac{bx}{a} + y \:=\:b$
Divide through by $\displaystyle b\!:\;\;\boxed{\frac{x}{a} + \frac{y}{b}\:=\:1}$