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Q) It is desired to construct a right triangle $ABC$ $(C=\pi/2) $ in the coordinate plane so that its legs are parallel to the $x$ and $y$ axes and so that the medians through A and B lie on the lines $y = 3x + 1$ and $y = mx + 2$. The number of different constants $m$ for which such a triangle exists is/are ?

I was able to identify a case where $m<3$ will only make a median but when i tried drawing a few more that was flawed . How do i begin an approach ?

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  • $\begingroup$ Check your question. Both medians through A and B should have negative slope if .the triangle's legs are parallel to the x and y axes. $\endgroup$ – SchrodingersCat Oct 31 '15 at 11:55
  • $\begingroup$ @Aniket yes but aren't there cases where slopes need not be negative ? $\endgroup$ – Sujith Sizon Oct 31 '15 at 12:01
  • $\begingroup$ Can you explain by a diagram? $\endgroup$ – SchrodingersCat Oct 31 '15 at 12:03
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    $\begingroup$ @Aniket a possible case (i think) i.stack.imgur.com/CG5hy.jpg $\endgroup$ – Sujith Sizon Oct 31 '15 at 12:07
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Let $A(s,3s+1),B(t,mt+2)$ where $m\not=3$.

Now let us separate it into cases.

Case 1 : $C(t,3s+1)$

Since the midpoint of the side $AC$ is on the line $y=mt+2$, $$3s+1=m\cdot\frac{s+t}{2}+2\iff (6-m)s-mt=2\tag1$$ Also, since the midpoint of the side $BC$ is on the line $y=3x+1$, $$\frac{3s+1+mt+2}{2}=3t+1\iff 3s+(m-6)t=-1\tag2$$ Eliminating $s$ from $(1)(2)$ gives $$(m-12)(m-3)t=12-m$$ Here, if $m\not=12$, then $s=t=\frac{1}{3-m}$, which is a contradiction.

So, in this case, the only possible $m$ is $m=12$. (it's easy to see that this is sufficient)

Case 2 : $C(s,mt+2)$

We have the followings : $$\frac{3s+1+mt+2}{2}=m\cdot s+2\iff (3-2m)s+mt=1\tag3$$ $$mt+2=3\frac{t+s}{2}+1\iff 3s+(3-2m)t=2\tag 4$$

Eliminating $s$ from $(3)(4)$ gives $$(4m-3)(m-3)t=-(4m-3)$$ If $4m-3\not=0$, then $t=s=\frac{1}{3-m}$, which is a contradiction.

So, in this case, the only possible $m$ is $m=3/4$.

Hence, the number of different constants $m$ for which such a triangle exists is $\color{red}{2}$.

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