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

& \Rightarrow x=\dfrac{\lambda (6)+2}{\lambda +1} \\\ & \Rightarrow x=\dfrac{6\lambda +2}{\lambda +1}.....(1) \\\ \end{aligned}$$ In the similar manner, $$\begin{aligned} & \Rightarrow y=\dfrac{\lambda (-2)+10}{\lambda +1} \\\ & \Rightarrow y=\dfrac{10-2\lambda }{\lambda +1}....(2) \\\ \end{aligned}$$ From equation (1) and equation (2), we get the coordinates of R as $$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{aligned} & \Rightarrow 6\lambda +2=5\left( \lambda +1 \right) \\\ & \Rightarrow \lambda =3....(3) \\\ \end{aligned}$$ So, it is clear that R (5,1) divides the line passing through the points P (2,10) and Q (6, -2) in the ratio $$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{aligned} & \Rightarrow X=\dfrac{3(6)-1(2)}{3-1} \\\ & \Rightarrow X=8....(4) \\\ \end{aligned}$$ In the similar manner, we get $$\begin{aligned} & \Rightarrow Y=\dfrac{(-2)(3)-10}{3-1} \\\ & \Rightarrow Y=-8....(5) \\\ \end{aligned}$$ So, the harmonic conjugate of point R (5,1) with respect to points P (2,10) and Q (6, -2) is A (8, -8). Note: Students should be careful while applying formulae. Some students may have a misconception that if two points $$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.