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Let $\mathrm{A}(2,3,5), \mathrm{B}(-1,3,2)$ and $\mathrm{C}(\lambda, 5, \mu)$ be the vertices of a $\triangle \mathrm{ABC}$. If the median through $\mathrm{A}$ is equally inclined to the coordinate axes, then:
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Verified Answer
The correct answer is:
$10 \lambda-7 \mu=0$
$10 \lambda-7 \mu=0$
If D be the mid-point of $B C$, then
$$
D=\left(\frac{\lambda-1}{2}, 4, \frac{\mu+2}{2}\right)
$$
Direction ratios of $\mathrm{AD}$ are $\frac{\lambda-5}{2}, 1, \frac{\mu-8}{2}$
Since median AD is equally inclined with coordinate axes, therefore direction ratios of $\mathrm{AD}$ will be equal, i.e,
$$
\begin{aligned}
&\frac{\left(\frac{\lambda-5}{2}\right)^2}{\left(\frac{\lambda-5}{2}\right)^2+1+\left(\frac{\mu-8}{2}\right)^2} \\
&=\frac{1}{\left(\frac{\lambda-5}{2}\right)^2+1+\left(\frac{\mu-8}{2}\right)^2} \\
&=\frac{\left(\frac{\mu-8}{2}\right)^2}{\left(\frac{\lambda-5}{2}\right)^2+1+\left(\frac{\mu-8}{2}\right)^2} \\
&\Rightarrow\left(\frac{\lambda-5}{2}\right)^2=1=\left(\frac{\mu-8}{2}\right)^2
\end{aligned}
$$
$$
\begin{aligned}
&\Rightarrow\left(\frac{\lambda-5}{2}\right)^2=1=\left(\frac{\mu-8}{2}\right)^2 \\
&\Rightarrow \lambda=7,3 \text { and } \mu=10,6
\end{aligned}
$$
If $\lambda=7$ and $\mu=10$
Then $\frac{\lambda}{\mu}=\frac{7}{10} \Rightarrow 10 \lambda-7 \mu=0$
$$
D=\left(\frac{\lambda-1}{2}, 4, \frac{\mu+2}{2}\right)
$$
Direction ratios of $\mathrm{AD}$ are $\frac{\lambda-5}{2}, 1, \frac{\mu-8}{2}$
Since median AD is equally inclined with coordinate axes, therefore direction ratios of $\mathrm{AD}$ will be equal, i.e,
$$
\begin{aligned}
&\frac{\left(\frac{\lambda-5}{2}\right)^2}{\left(\frac{\lambda-5}{2}\right)^2+1+\left(\frac{\mu-8}{2}\right)^2} \\
&=\frac{1}{\left(\frac{\lambda-5}{2}\right)^2+1+\left(\frac{\mu-8}{2}\right)^2} \\
&=\frac{\left(\frac{\mu-8}{2}\right)^2}{\left(\frac{\lambda-5}{2}\right)^2+1+\left(\frac{\mu-8}{2}\right)^2} \\
&\Rightarrow\left(\frac{\lambda-5}{2}\right)^2=1=\left(\frac{\mu-8}{2}\right)^2
\end{aligned}
$$
$$
\begin{aligned}
&\Rightarrow\left(\frac{\lambda-5}{2}\right)^2=1=\left(\frac{\mu-8}{2}\right)^2 \\
&\Rightarrow \lambda=7,3 \text { and } \mu=10,6
\end{aligned}
$$
If $\lambda=7$ and $\mu=10$
Then $\frac{\lambda}{\mu}=\frac{7}{10} \Rightarrow 10 \lambda-7 \mu=0$
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