Find the equation of a line given the bounded area

Question

Find the equation of the line through $(2, 2)$ and forming with the axes a triangle of area $9$.

Answer 1

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 $$\dfrac x6+\dfrac y3=1\;and\;\dfrac x3+\dfrac y6=1$$

Answer 2

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 ;-)

Answer 3

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 $m>0$