How to prove that $\frac{x}{a} + \frac{y}{b} = 1$ where $a$ is $x$-intercept and $b$ is $y$-intercept

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.

Solution 1:

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 $\frac yb + \frac xa =1$

Solution 2:

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 $$\frac xa + \frac yb = 1$$

Solution 3:

$x$-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.