Question
It is desired to construct a right angled triangle $ABC(\angle C = \pi/2)$ in $xy$-plane so that its sides are parallel to co-ordinates axes and the medians through $A$ and $B$ lie on the lines $y = 3x + 1$ and $y = mx + 2$ respectively. The values of ' $m$ ' for which such a triangle is possible is/are :
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Solution
Sol. [Key Concept: Right triangle with axis-parallel sides and medians on given lines ⇒ slope condition fixes $m$.]
$$tan\theta_{1} = \frac{BC}{2AC} $$
$$\begin{matrix} tan\theta_{2} & \ = \frac{BC}{(AC/2)} = \frac{2BC}{AC} \\ \frac{tan\theta_{1}}{tan\theta_{2}} & \ = \frac{1}{4} \end{matrix}$$
Case-I : If $m < 3$,
$$\begin{matrix} tan\theta_{1} & \ = \frac{1}{3}\text{~}\text{and}\text{~}tan\theta_{2} = \frac{1}{m} \\ \frac{m}{3} & \ = \frac{1}{4} \Rightarrow m = \frac{3}{4} \end{matrix}$$
Case-II : If $\ m > 3$,
$$\begin{matrix} tan\theta_{1} & \ = \frac{1}{m}\text{~}\text{and}\text{~}tan\theta_{2} = \frac{1}{3} \\ \Rightarrow \ \frac{3}{m} & \ = \frac{1}{4} \Rightarrow m = 12 \end{matrix}$$
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