Question

Find the harmonic conjugate of point R (5,1) with respect to points P (2,10) and Q (6, -2).

Answer 1

Hint: We will assume that R (5, 1) divides the line passing through P (2,10) and Q (6, -2) in the ratio\[\lambda :1\] internally. We know that if two points \[A({{x}_{1}},{{y}_{1}})\] and \[B({{x}_{2}},{{y}_{2}})\] are divided by the point \[C({{x}_{3}},{{y}_{3}})\] in the ratio \[m:n\] internally, then we get \[{{x}_{3}}=\dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n}\] and \[{{y}_{3}}=\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n}\]. We were given the coordinates of R (5,1). By this we can get a relation between \[\lambda \] and coordinates of R. By this we can find the value of \[\lambda \]. Now to find the harmonic conjugate point we have to find a point which divides the line passing through P (2,10) and Q (6, -2) in the ratio \[\lambda :1\] externally. We know that if two points \[A({{x}_{1}},{{y}_{1}})\] and \[B({{x}_{2}},{{y}_{2}})\] are divided by the point \[C({{x}_{3}},{{y}_{3}})\] in the ratio \[m:n\] externally, then we get \[{{x}_{3}}=\dfrac{m{{x}_{2}}-n{{x}_{1}}}{m-n}\] and \[{{y}_{3}}=\dfrac{m{{y}_{2}}-n{{y}_{1}}}{m-n}\]. By using the value of \[\lambda \], we can find the coordinates of harmonic conjugate point if R (5,1) with respect to points P (2,10) and Q (6, -2).

Complete step-by-step answer:
Let us assume that R (5,1) divides the line passing through the points P (2,10) and Q (6, -2) in the ratio \[\lambda :1\] internally.
So, now we should find the coordinates of a point R in terms of \[\lambda \].
We know that if two points \[A({{x}_{1}},{{y}_{1}})\] and \[B({{x}_{2}},{{y}_{2}})\] are divided by the point \[C({{x}_{3}},{{y}_{3}})\] in the ratio \[m:n\] internally, then we get \[{{x}_{3}}=\dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n}\] and \[{{y}_{3}}=\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n}\].
 So, if two points P (2,10) and Q (6, -2) are divided by a point \[R(x,y)\] in the ratio \[\lambda :1\] internally.
Then we get
\[\begin{align}
  & \Rightarrow x=\dfrac{\lambda (6)+2}{\lambda +1} \\
 & \Rightarrow x=\dfrac{6\lambda +2}{\lambda +1}.....(1) \\
\end{align}\]
In the similar manner,
\[\begin{align}
  & \Rightarrow y=\dfrac{\lambda (-2)+10}{\lambda +1} \\
 & \Rightarrow y=\dfrac{10-2\lambda }{\lambda +1}....(2) \\
\end{align}\]
From equation (1) and equation (2), we get the coordinates of R as \[R\left( \dfrac{6\lambda +2}{\lambda +1},\dfrac{10-2\lambda }{\lambda +1} \right)\].
We know that the coordinates of R as R (5,1).
So,
\[\dfrac{6\lambda +2}{\lambda +1}=5\]
By using cross multiplication, we get
\[\begin{align}
  & \Rightarrow 6\lambda +2=5\left( \lambda +1 \right) \\
 & \Rightarrow \lambda =3....(3) \\
\end{align}\]
So, it is clear that R (5,1) divides the line passing through the points P (2,10) and Q (6, -2) in the ratio \[3:1\] internally.
Now to find harmonic conjugate point of point R (5,1) with respect to points P (2,10) and Q (6, -2), we need to find a point which divides the line P (2,10) and Q (6,-2) in the ratio \[3:1\] externally.
Let the harmonic point of R (5,1) is \[A(X,Y)\].
We know that if two points \[A({{x}_{1}},{{y}_{1}})\] and \[B({{x}_{2}},{{y}_{2}})\] are divided by the point \[C({{x}_{3}},{{y}_{3}})\] in the ratio \[m:n\] externally, then we get \[{{x}_{3}}=\dfrac{m{{x}_{2}}-n{{x}_{1}}}{m-n}\] and \[{{y}_{3}}=\dfrac{m{{y}_{2}}-n{{y}_{1}}}{m-n}\].
So, we get
\[\begin{align}
  & \Rightarrow X=\dfrac{3(6)-1(2)}{3-1} \\
 & \Rightarrow X=8....(4) \\
\end{align}\]
In the similar manner, we get
\[\begin{align}
  & \Rightarrow Y=\dfrac{(-2)(3)-10}{3-1} \\
 & \Rightarrow Y=-8....(5) \\
\end{align}\]
So, the harmonic conjugate of point R (5,1) with respect to points P (2,10) and Q (6, -2) is A (8, -8).

Note: Students should be careful while applying formulae. Some students may have a misconception that if two points \[A({{x}_{1}},{{y}_{1}})\] and \[B({{x}_{2}},{{y}_{2}})\] are divided by the point \[C({{x}_{3}},{{y}_{3}})\] in the ratio \[m:n\] internally, then we get \[{{x}_{3}}=\dfrac{m{{x}_{2}}-n{{x}_{1}}}{m-n}\] and \[{{y}_{3}}=\dfrac{m{{y}_{2}}-n{{y}_{1}}}{m-n}\] . And they may also have a misconception that if two points \[A({{x}_{1}},{{y}_{1}})\] and \[B({{x}_{2}},{{y}_{2}})\] are divided by the point \[C({{x}_{3}},{{y}_{3}})\] in the ratio \[m:n\] internally, then we get \[{{x}_{3}}=\dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n}\] and \[{{y}_{3}}=\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n}\]. This should be avoided to solve this problem accurately.