Question

It is desired to construct a right angled triangle ABC $(C=\pi /2)$ in xy plane so that its sides are parallel to coordinates axes and the medians through A and B lie on the lines $y=3x+1$ and $y=mx+2$ respectively. The value of $m$ for which such a triangle is possible is /are:

• A. $12$
• B. $\frac{3}{4}$
• C. $\frac{4}{3}$
• D. $\frac{1}{12}$

Answer 1

The correct option is B $\frac{3}{4}$

Let $A(s,3s+1),B(t,mt+2)$

Now

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

Since the midpoint of side AC is on line $y=mx+2$,

$3s+1=m(\frac{s+t}{2}+2)\Rightarrow (6-m)s-mt=2----\left(1\right)$

Also,since the midpoint of the side BC is on line $y=3x+1$

$\frac{3s+1+mt+2}{2}=3t+1\Rightarrow 3s+(m-6)t=-1---\left(2\right)$

Eliminating s from (1) and (2)

$(m-12)(m-3)t=12-m$

Here, if $m\ne 12$ then $s=t=\frac{1}{3-m}$, which is contradiction

So,in this case, the only possible m is $m=12$

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

We have the followibg

$\frac{3s+1+mt+2}{2}=ms+2\Rightarrow (3-2m)s+mt=1-----\left(3\right)$

$mt+2=3\frac{t+s}{2}+1\Rightarrow 3s+(3-2m)t=2-------\left(4\right)$

Eliminating s from (3) and (4) we get

$(4m-3)(m-3)t=-(4m-3)$

If $4m-3=0$

$m=\frac{3}{4}$