Find the equation of a line given the bounded area Find the equation of the line through $(2, 2)$ and forming with the axes a triangle of area $9$.
 A: suppose eqn of desirable line is$$\dfrac xa+\dfrac yb=1$$
since it is making triangle with axes,so area of triangle are
$$\dfrac 12 ab=9\implies ab=18$$
$$\dfrac xa+\dfrac yb=1$$
since line is passing by ($2,2$)
$$\dfrac 2a+\dfrac 2b=1$$
$$2a+2b=ab$$
$$a+b=9$$
so we've equations $a+b=9\;$and $ab=18$
$$ab=18\implies b=\dfrac {18}{a}$$
$$a+b=9$$
$$a+\dfrac {18}{a}=9\implies a^2-9a+18=0\implies a=6,3\; and\;b=3,6$$
so equation of lines are $$\dfrac x6+\dfrac y3=1\;and\;\dfrac x3+\dfrac y6=1$$
A: There are 4 solutions (this should be a comment really, answer because of picture).
$$10.68 \simeq \frac{3\sqrt{17}+9}{2}$$

I hope this helps ;-)
A: The equation of any line passing through $(2,2)$ can be written as $$\frac{y-2}{x-2}=m$$ where $m$ is the gradient.
Rearranging in  Intercept form, $$\frac x{\frac{2m-2}m}+\frac y{-(2m-2)}=1$$
So, the area of the Triangle will be $$=\frac12\left|\frac{(2m-2)^2}{-m}\right|=\frac{2m^2-4m+2}{|m|}$$
Clearly, $m\ne0$
If $m<0,$  $$-\frac{2m^2-4m+2}m=9\iff 2m^2+5m+2=0$$
$$m=\frac{-5\pm\sqrt{5^2-4\cdot2\cdot2}}{2\cdot2}=\frac{-5\pm3}4=-2,-\frac12$$
Similarly, for $m>0$
