Answer
$(a)y=-\frac{b}{a}x+b$
(b) x-intercept
$x=a$
y-intercept
$y=b$
(c) Intercepts $x=a $ and $y=b$
Work Step by Step
$\frac{x}{a}+\frac{y}{b}=1$
(a)
$\frac{x}{a}+\frac{y}{b}=1$
Multiplying with ab
$[\frac{x}{a}+\frac{y}{b}]\times ab=1\times ab$
$bx+ay=ab$
$ay=-bx+ab$
$\frac{ay}{a}=\frac{bx+ab}{a}$
$y=-\frac{b}{a}x+b$
(b)
Putting $y=0$ in
$\frac{x}{a}+\frac{y}{b}=1$
$\frac{x}{a}+\frac{0}{b}=1$
$\frac{x}{a}=1$
x-intercept
$x=a$
Putting $x=0$ in
$\frac{x}{a}+\frac{y}{b}=1$
$\frac{0}{a}+\frac{y}{b}=1$
$0+\frac{y}{b}=1$
$\frac{y}{b}=1$
y-intercept
$y=b$
(c)
From the equation
$y=-\frac{b}{a}x+b$
y-intercept is b
At y=0
$0=-\frac{b}{a}x+b$
$-\frac{b}{a}x+b=0$
$-\frac{b}{a}x=-b$
$x=a$ is x-intercept
Hence
$\frac{x}{a}+\frac{y}{b}=1$ is intercept form of a line
Moreover
$y=0$ $\Longrightarrow$ $x=a$ is x-intercept
Similarly
$x=0$ $\Longrightarrow$ $y=b$ is y-intercpt
S0
$\frac{x}{a}+\frac{y}{b}=1$ is in intercept form
