The harmonic conjugate of $(2,3,4)$ with respect to the points $(3,-2,2),(6,-17,-4)$ is

The harmonic conjugate of $(2,3,4)$ with respect to the points $(3,-2,2),(6,-17,-4)$ is
  1. $\left(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}\right)$
  2. $\left(\frac{18}{5}, -5, \frac{4}{5}\right)$
  3. $\left(\frac{-18}{5},\frac{5}{4}, \frac{4}{5}\right)$
  4. $\left(\frac{18}{5}, -5, \frac{-4}{5}\right)$

Solution

Let we points $\mathrm{P}(2,3,4), \mathrm{A}(3,-2,2)$ and $\mathrm{B}(6,-17,-4)$. and $\mathrm{P}$ divides $\mathrm{AB}$ in the ratio $k: 1$ then $\begin{aligned} & (2,3,4)=\left(\frac{6 k+3}{k+1}, \frac{-17 k-2}{k+1}, \frac{-4 k+2}{k+1}\right) \\ & 2=\frac{6 k+3}{k+1} \\ & \Rightarrow \quad 2 k+2=6 k+3 \\ & \Rightarrow-4 k=1 \Rightarrow k=-\frac{1}{4} \end{aligned}$ Harmonic conjugate $\mathrm{Q}$ divides in the ratio $-k: 1$, So Ratio will be $\frac{1}{4}: 1$ $\therefore \quad$ Coordinates of $\mathrm{Q}$ $\begin{aligned} & =\left(\frac{\frac{1}{4}(6)+3}{\frac{1}{4}+1}, \frac{\frac{1}{4}(-17)-2}{\frac{1}{4}+1}, \frac{\frac{1}{4}(-4)+2}{\frac{1}{4}+1}\right) \\ & =\left(\frac{6+12}{5}, \frac{-17-8}{5}, \frac{-4+8}{5}\right) \\ & =\left(\frac{18}{5},-5 \frac{4}{5}\right) \end{aligned}$

Asked in: AP EAMCET 2016

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