Exam

JEE Main

Year

2014

Date

11 April

Mode

Online

Subject

Mathematics

Chapter

Properties of Triangles

JEE Main 2014 · 11 April · Online · Mathematics · Properties of Triangles

Let A(2,3,5), B(-1,3,2) and C(λ, 5, μ) be the vertices of a ABC ....

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:

Step-by-step solution

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$10 \lambda-7 \mu=0$