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.

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- Jan 16th 2007, 01:51 PMsymmetryEquation 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. - Jan 16th 2007, 02:15 PMtopsquark
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 - Jan 16th 2007, 05:18 PMsymmetryok
So, I simply needed to set y = 0 to find the x-intercepts and set x = 0 to find the y-intercepts.

How about that?

Thanks! - Jan 16th 2007, 05:45 PMtopsquark
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 - Jan 16th 2007, 06:49 PMSoroban
Hello, symmetry!

You may be expect to*derive*that formula . . .

Quote:

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