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In an isosceles triangle $\mathrm{ABC}$ , the vertex $\mathrm{A}$ is $(6,1)$ and the equation of the base $\mathrm{BC}$ is $2 x+y=4$ . Let the point $\mathrm{B}$ lie on the line $x+3 y=7.$ If $(\alpha, \beta)$ is the centroid of $\Delta \mathrm{ABC}$ , then $15(\alpha+\beta)$ is equal to :
Option: 1
$39$

Option: 2
$41$

Option: 3
$51$

Option: 4
$63$

Answers (1)

best_answer

Foot of perpendicular from $\mathrm{A(6,1) \text { to } B C}$

$\mathrm{\frac{x-6}{2}=\frac{y-1}{1}=\frac{-1\left(6 x^{2}+1-4\right)}{2^{2}+1^{2}}=-\frac{9}{5}} \\$

$\mathrm{\Rightarrow x=\frac{12}{5}, y=\frac{-4}{5}}$

Centroid divide medium in $\mathrm{2:1}$

$\mathrm{\frac{A\left ( 6,1 \right )G\left ( \alpha ,\beta \right )}{2:1}D\left ( \frac{12}{5},\frac{-4}{5} \right )}$

$\mathrm{\text { So } \alpha=\frac{6 \times 1+\frac{12}{5} \times 2}{2+1}, \quad \beta=\frac{1 \times 1+2 \times\left(-\frac{4}{5}\right)}{2+1}}$

$\mathrm{\alpha+\beta=\frac{30+24+5-8}{15}=\frac{51}{15} \Rightarrow 15(\alpha+\beta)=51}$

Hence the correct answer is option 3.

Posted by

Sanket Gandhi

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