The triangle ABC is right angled at C, has medians AD,BE,CF. AD lies along the line $y = x + 3$ , BE lies along the line $y = 2 x + 4$ . If the length of the hypotenuse is 60, then area of the triangle ABC in square units is……………

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answer is 400.

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Detailed Solution

Area = $\frac{1}{2} a b$
Given: $(- 1, 2) & " α "$ be acute angle between the medians
$\tan α = \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} = \frac{2 - 1}{1 + 2} = \frac{1}{3}$

Now, $π - α + 90^{0} + θ + β = 360$

$\begin{array}{l} α = θ + β - 90^{0} \\ \cot α = - \tan (θ + β) \\ - 3 = \frac{+ \tan θ + \tan β}{1 - \tan θ \tan β} \\ = \frac{\frac{2 b}{a} + \frac{2 a}{b}}{1 - \frac{2 b}{a}, \frac{2 a}{b}} \\ 9 = \frac{2 (a^{2} + b^{2})}{a b} \\ \Rightarrow a b = \frac{2 \times 3600}{9} \\ \Rightarrow \frac{1}{2} a b = 400 \end{array}$