Question

Find equation of the line passing through the point $(2,2)$ and cutting off intercepts on the axes whose sum is $9$.

Answer 1

The equation of a line in the intercept form is
$\frac{x}{a}+\frac{y}{b}=1$.......(1)
where $a$ and $b$ are intercepts on $x$ and $y$ axes respectively.
it is given that $a+b=9\Rightarrow b=9-a$.......(2)
From equation (1), we obtain
$\frac{x}{a}+\frac{y}{9-a}=1$
Also it is passing through $(2,2)$
$\frac{2}{a}+\frac{2}{9-a}=1$
$\Rightarrow 2\{\frac{1}{a}+\frac{1}{9-a}\}=1$
$\Rightarrow 2\left\{\frac{9-a+a}{a(9-a)}\right\}=1$
$\Rightarrow \frac{18}{9a-a^2}=1$
$\Rightarrow 18=9a-a^2$
$\Rightarrow a^2-6a-3a+18=0$
$\Rightarrow a(a-6)-3(a-6)=0$
$\Rightarrow (a-6)(a-3)=0$
$\Rightarrow a=6$ or $a=3$
If $a=3\Rightarrow b=9-3=6$, then the equation of the line is
$\frac{x}{3}+\frac{y}{6}=1\Rightarrow 2x+y-6=0$
And if $a=6\Rightarrow b=9-6=3$, then the equation of the line is
$\frac{x}{6}+\frac{y}{3}=1\Rightarrow x+2y-6=0$