Question

The harmonic conjugate of $(4,1)$ with respect to the points $(3,2)$ and$(-1,6)$ is

Answer 1

Hint: Approach the solution by applying the section formula for given points. Here is the section formula for x coordinate and the section formula for y coordinates is similar as x coordinate.
Section formula $x={\displaystyle \frac{m{x}_2+n{x}_1}{m+n}}$

Here we have to find harmonic conjugate of $(4,1)$ with respect to given points
Let $(4,1)$ divides $(3,2)$ and $(-1,6)$ in $K:1$ ratio
So here let us apply the section formula
Section formula $x={\displaystyle \frac{m{x}_2+n{x}_1}{m+n}}$
$$\begin{gathered} \Rightarrow 4={\displaystyle \frac{k(-1)+1(3)}{k+1}} \\ \Rightarrow 4k+4=3-k \\ \Rightarrow 5k=-1 \\ \Rightarrow k={\displaystyle \frac{-1}{5}} \end{gathered}$$
So, here the given points $(3,2)$and $(-1,6)$ are going divide in $-1:5$ ratio
Here the ratio $-1:5$ divides the points externally but we have to divide the ratio internally
So to get the internal point ratio we have to remove the negative sign from the external ratio.
$\therefore$ Internal ratio =$1:5$
The harmonic conjugate divides the given point internally in ratio $1:5$
Apply the section formula
$x={\displaystyle \frac{m{x}_2+n{x}_1}{m+n}}$
$$\begin{gathered} \Rightarrow {\displaystyle \frac{1(-1)+5(3)}{5+1}} \\ \Rightarrow {\displaystyle \frac{7}{3}} \end{gathered}$$
$$\begin{gathered} y={\displaystyle \frac{m{y}_2+n{y}_1}{m+n}} \\ \Rightarrow y={\displaystyle \frac{1(6)+5(2)}{5+1}} \\ \Rightarrow y={\displaystyle \frac{8}{3}} \end{gathered}$$
Therefore the harmonic conjugate of the required point that divides internally in the ratio $1:5$ = $\left({\displaystyle \frac{8}{3}},{\displaystyle \frac{7}{3}}\right)$

Note: In these types of problems external or internal ratio matter where sign value is different. Here we have used section formulas to both x and y coordinates.