Question

Find the harmonic conjugate of (4,1) with respect to the points (3,2) and (-1,6).

Answer 1

The harmonic conjugate is $(\frac{7}{3}, \frac{8}{3})$.

Answer 2

To find the harmonic conjugate of a point P with respect to two other points A and B, we use the concept of section formula and the definition of harmonic range.

Let the given points be: P = (4,1) A = (3,2) B = (-1,6)

Let Q = (x,y) be the harmonic conjugate of P with respect to A and B.

1. Determine the ratio in which P divides the segment AB. Let P divide AB in the ratio $\lambda:1$. Using the section formula: $P = \left( \frac{\lambda x_B + 1 x_A}{\lambda + 1}, \frac{\lambda y_B + 1 y_A}{\lambda + 1} \right)$

Substitute the coordinates of P, A, and B: For the x-coordinate: $4 = \frac{\lambda(-1) + 1(3)}{\lambda + 1}$ $4(\lambda + 1) = -\lambda + 3$ $4\lambda + 4 = -\lambda + 3$ $5\lambda = -1$ $\lambda = -\frac{1}{5}$

The negative value of $\lambda$ indicates that P divides the segment AB externally in the ratio $1:5$. This means the ratio of directed distances $\frac{\vec{AP}}{\vec{PB}} = -\frac{1}{5}$.

2. Find the harmonic conjugate Q. By definition, if P divides AB in the ratio $\lambda$, then its harmonic conjugate Q divides AB in the ratio $-\lambda$. So, Q divides AB in the ratio $- (-\frac{1}{5}) = \frac{1}{5}$. This positive ratio indicates that Q divides the segment AB internally in the ratio $1:5$.

Now, use the section formula for internal division to find the coordinates of Q: $Q = \left( \frac{1 \cdot x_B + 5 \cdot x_A}{1 + 5}, \frac{1 \cdot y_B + 5 \cdot y_A}{1 + 5} \right)$

Substitute the coordinates of A and B: $Q_x = \frac{1(-1) + 5(3)}{6} = \frac{-1 + 15}{6} = \frac{14}{6} = \frac{7}{3}$ $Q_y = \frac{1(6) + 5(2)}{6} = \frac{6 + 10}{6} = \frac{16}{6} = \frac{8}{3}$

Therefore, the harmonic conjugate of (4,1) with respect to (3,2) and (-1,6) is $(\frac{7}{3}, \frac{8}{3})$.