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Two variables $x$ and $y$ are related by a certain equation. This equation may be expressed in two forms suitable for drawing straight line graphs. The two graphs shown have different variables plotted at each axis. Given the coordinates of a point on each line, find the original equation relating $x$ and $y$.

I am completely stuck and do not know how to proceed.Please refer to this for the two graphs

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From the second graph we have,

$$x=m\frac{y}{x}+b$$

This is useful be we can solve for $y$. To get,

$$\frac{x^2-bx}{m}=y$$

So $y$ is a quadratic in $x$, we also have from the second graph that one point on our quadratic is given by,

$$x=6$$

$$\frac{y}{x}=2 \implies y=6(2)=12$$

So one point on our quadratic is $(6,12)$.

From our first graph we have,

$$\frac{x}{y}=1 \implies x=y$$

$$\frac{x^2}{y}=11 \implies x=11=y$$

So another point on the quadratic is $(11,11)$, then substituting our coordinates into $x=m\frac{y}{x}+b$ we have the system,

$$11=m+b$$

$$6=2m+b$$

Whose solution is $m=-5$ and $b=16$ so that,

$$y=-\frac{x^2-16x}{5}$$

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    Wait, I don't understand... what do you mean find two other points? Do you mean use the two points that were given in the diagrams? – 2017-02-11
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    What is the point of finding out that the origin is a point on the quadratic? – 2017-02-11
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    No I mean $(x,y)$. I wrote that to emphasize we have $3$ points that's enough to explicitly define a quadratic. @SydneySniper – 2017-02-11
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    I still don't understand...can you give the next 2 steps? I think I can work form that... – 2017-02-11
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    Actually, it's fine I solved it by comparing the two equations – 2017-02-11