Question

Harmonic conjugate of the point C $(5,1)$ with respect to the point A $(2,10)$ and B $(6, - 2)$ is?

Answer 1

Initially we need to find in which ratio the given point C is dividing the line joining the given points A and B in our problem. Once we find that, we need to deal with the ratio as the harmonic conjugate point will divide it externally. Thus, solving these we will find our desired answer.

Complete step by step solution:

Let, C $(5,1)$ is harmonic conjugate with respect to A $(2,10)$ and B $(6, - 2)$
$A = \left( {2,10} \right),B = \left( {6, - 2} \right),C = \left( {5,1} \right)$
Let the ratio of which C divides AB is $\lambda :1$ ,
Now, if we find a point C dividing a line AB, $m:n$, and if A and B has coordinates $(a,b)$ and $(c,d)$ respectively, then, C has a coordinate, $(\dfrac{{an + cm}}{{m + n}},\dfrac{{bn + dm}}{{m + n}})$ .
Then we have here, $m = \lambda ,n = 1$ and coordinate of A and B are given,
So, $\left( {\dfrac{{2 + 6\lambda }}{{\lambda + 1}}\,,\dfrac{{ - 2 + 10\lambda }}{{\lambda + 1}}} \right)\, = (5,1\;)$
$\Rightarrow$ $\dfrac{{2 + 6\lambda }}{{\lambda + 1}}\,\, = 5\;$and $\dfrac{{ - 2 + 10\lambda }}{{\lambda + 1}}\, = 1$
From, $\dfrac{{2 + 6\lambda }}{{\lambda + 1}}\,\, = 5\;$
On cross multiplication we get,
$\Rightarrow 2 + 6\lambda = 5(\lambda + 1)\;$
On simplification we get,
$\Rightarrow 2 + 6\lambda = 5\lambda + 5$
Changing sides, we get,
$\Rightarrow 6\lambda - 5\lambda = 5 - 2$
On solving we get $\lambda = 3$
So, we get the ratio as, $3:1$ , which divides externally in case of the harmonic conjugate.
Now, if we find a point C dividing a line AB, $m:n$, and if A and B has coordinates $(a,b)$ and $(c,d)$ respectively, then, C has a coordinate, $(\dfrac{{an - cm}}{{m - n}},\dfrac{{bn - dm}}{{m - n}})$.
Then we have our desired point as,
$= \left( {\dfrac{{3 \times 6 - 1 \times 2}}{{3 - 1}},\dfrac{{3( - 2) - 1 \times 10}}{{3 - 1}}} \right)$
$= \left( {\dfrac{{18 - 2}}{2},\dfrac{{ - 6 - 10}}{2}} \right)$
$= (8, - 8)$
Hence the conjugate of point C $(5,1)$ is $(8, - 8)$.

Note: The harmonic conjugate point of an ordered triple of points on the real projective line is defined by the following construction:
Given three collinear points A, B, C, let L be a point not lying on their join and let any line through C meet LA, LB at M, N respectively. If AN and BM meet at K, and LK meets AB at D, then D is called the harmonic conjugate of C with respect to A, B.