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How to prove that $\dfrac{x}{a} + \dfrac{y}{b} = 1\;$ where $\,a\,$ is the $\,x$-intercept and $\,b\,$ is the $\,y$-intercept for all $\,a,b \neq 0$

This was a question on my son's math analysis test today, and neither of us is sure how to approach solving it.

Daniel W. Farlow
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user62071
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    Intercepts of what line/curve? – user7530 Feb 13 '13 at 00:13
  • The formulation of the problem is pretty bad; it should at least say that the subject is a straight line. Other kinds of curves with the given intercepts would have very different equations. I think a reasonable formulation would be "Prove that $\frac xa+\frac yb=1$ is the equation of the straight line with $x$-intercept $a$ and $y$-intercept $b$ whenever $a,b\neq0$." – Andreas Blass Feb 13 '13 at 15:12
  • @AndreasBlass, I apologize for not being clearer. We were discussing the problem on the way home from school and I entered it on this site from memory, omitting, as you point out, that the equation describes a straight line. – user62071 Feb 13 '13 at 20:38

6 Answers6

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If the x intercept is $a$ then $(a,0)$ is on the line. If y-intercept is b then $(0,b)$ is on the line. Now you have two points of the line and can get its formula. The slope between thse points is $m={(0-b)\over(a-0)}=\frac{-b}{a}$. So the equation of line is $y=mx+b=\frac{-bx}{a}+b$. Now rearrange as $y+\frac{bx}{a}=b$ divide by $b$ to get $\frac yb + \frac xa =1$

AltercatingCurrent
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Maesumi
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Equation of the line is $$ Ax + By + C = 0 $$ If x-interception is $M(a,0)$ and y-interception is $N(0, b)$ then they need to satisfy line equation $$ Aa + C = 0;\qquad Bb + C = 0 $$ from which you can find $$ A = -\frac Ca; \qquad B = -\frac Cb $$ After substituting to line equation $$ -\frac Cax - \frac Cby = -C $$ or $$\frac xa + \frac yb = 1$$

Kaster
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    (+1): It's worth noting that for $Ax+By+C=0$ to be an equation for a line, we must have $A\neq 0$ or $B\neq 0$. From this, and the work you did, it follows in particular that $C\neq 0$, which allows us to take the last step. – Cameron Buie Feb 13 '13 at 00:32
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$x$-intercept $= a\implies y = 0 \implies (a,0)$ is on the given line.

$y$-intercept $ = b \implies x = 0 \implies (0,b)$ is on the given line.

Now you have two points of the line use those points to find slope. The slope of the line determined by these points is $$m = \dfrac{y_1 - y_2}{x_1 - x_2} =\dfrac {0-b}{a-0}=-\dfrac ba$$ Using the slope-intercept equation of line:$$y=mx+b$$ and substituting slope $m =- \dfrac ba$ gives us: $$y = -\dfrac ba x +b\tag{1}$$ Rearranging gives us

$$y+\cfrac ba = b$$ Now divide by $b$ to get the desired form of the equation: $$\dfrac yb + \dfrac xa =1\tag{2}$$

Note that equations $(1)$ and $(2)$ represent the same exact line; they are simply two forms in which we can express that line.

amWhy
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You can think of it also in geometrical terms. If you scale the $x$-axis by dividing by $a$ and the y-axis by dividing by $b$, then in your new coordinate system, with coordinates $x'=x/a$ and $y'=y/b$ the line will obviously go through the points $(0,1)$ and $(1,0)$. The equation for this line $y'=x'-1$ or $x'+y'=1$. Substitute the definitions of $x'$ and $y'$ and you're done.

(NB: It becomes even more obvious, if you draw that as a diagram. Unfortunately I don't know how to do this on SE.)

Elmar Zander
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Proof by demonstration:

$$\begin{align} (x,y) = (a,0): &\qquad \frac{x}{a} + \frac{y}{b} = \frac{a}{a} + \frac{0}{b} = 1 + 0 = 1 \\[8pt] (x,y) = (0,b): &\qquad \frac{x}{a} + \frac{y}{b} = \frac{0}{a} + \frac{b}{b} = 0 + 1 = 1 \end{align}$$

A linear equation ---that is, an equation of the form $Ax+By+C=0$ with $A$ and $B$ not both zero--- satisfied by the coordinates of two distinct points on a line is an equation of that line. (To paraphrase Euclid: "Two points determine a line equation.") The nature of the points used says exactly that the line's $x$-intercept is "$a$" and that its $y$-intercept is "$b$". And there you go.

It's interesting that a couple of answers leverage the slope-intercept form of the line equation to "prove" the intercept-intercept form. Certainly, slope-intercept is more familiar than intercept-intercept, but I submit that it's actually less obvious, so that its use actually complicates matters (if only slightly).

Blue
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Intercept form : If $a$ and $b$ are the intercepts made by a line on the axes of $x$ and $y$, its equation is written $$\frac{x}{a} + \frac{y}{b} = 1$$

enter image description here

To find the equation of $\overleftrightarrow{l}$ in intercept form, let’s first find the equation of $\overleftrightarrow{l}$ in slope intercept form. Observe that $\overleftrightarrow{l}$ passes through the point $(a,0)$ and intersects the y-axis at $(0,b)$. This means the

  • slope $(m)$ of $\overleftrightarrow{l}$ is $-\dfrac{b}{a}$, and
  • the y-intercept of $\overleftrightarrow{l}$ is $b$

So, we can write the equation of $\overleftrightarrow{l}$ in slope-intercept form as $$ y= -\dfrac{b}{a} x+b$$

Now, by adding the term $\dfrac{b}{a} x$ on both sides of this equation, we can rewrite it as \begin{aligned}y+\dfrac{b}{a}x=b\\ \Rightarrow \dfrac{y}{b}+\dfrac{x}{a}=1\end{aligned} [Dividing both LHS and RHS by $b$.]

This is known as the equation of a line in intercept form, where $a$ and $b$ are $x$ and y-intercepts of the line respectively as shown below

enter image description here

Practice question:

What is the equation of the line with $5$ and $-3$ as $x-$ and y-intercept, respectively? See the Solution.