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

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closed as off-topic by Thomas, 1999, jkabrg, Did, quid Jul 9 '15 at 23:15

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i guess it's intercept method, but i'm not sure. – mona Jul 15 '13 at 10:26

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

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@dtldarek 4? I can only calculate $2$. I certainly do not think there are an infinite number of lines. – Daryl Jul 15 '13 at 12:13
sorry guys my mistake I forget that there are two equations.I've corrected it now – iostream007 Jul 15 '13 at 12:16
@Daryl Yep, 4 solutions, there is a second pair of equations, where $ab=-18$, those would be triangles in II and IV quadrants. – dtldarek Jul 15 '13 at 12:33
@dtldarek but if you take ab=-18 case then those line will not passing by from (2,2) – iostream007 Jul 15 '13 at 12:35
Yes, they will, though of course the given point will not lie between two of the triangle's vertices (as with the quadrant 1 cases). Play with it. You'll get it. More generally, we want $$\frac2a+\frac2b=1$$ and $$\frac12|a||b|=9.$$ The first of these equations prevents $a,b$ from both being negative (neither can be $0$), and you've taken care of the cases that $a,b>0$. Only the $a<0<b$ and $b<0<a$ cases remain, both of which make the latter equation $$-\frac12ab=9,$$ or $$ab=-18.$$ – Cameron Buie Jul 15 '13 at 12:47

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 ;-)

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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 $m<0,$ $$-\frac{2m^2-4m+2}m=9\iff 2m^2+5m+2=0$$


Similarly, for $m>0$

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