Midpoint of Segment Joining Ends of Chords Bisected by Y-Axis

To find the value of $6(\alpha + \beta)$, we begin by analyzing the geometry of the given circle and chords.

The equation of the given circle is: $C: x^2 + y^2 + x - 3y = 0$

Let $P(1, 2)$ be the given point on the circle. A chord is drawn from $P(1, 2)$ and is bisected by the $y$-axis.

Since the $y$-axis bisects the chord, the midpoint of the chord must lie on the $y$-axis. Thus, we can represent the midpoint as $M(0, k)$ for some real number $k$.

Let $R(x_1, y_1)$ be the other endpoint of the chord. Since $M(0, k)$ is the midpoint of the segment joining $P(1, 2)$ and $R(x_1, y_1)$, we use the midpoint formula: $\left( \frac{1 + x_1}{2}, \frac{2 + y_1}{2} \right) = (0, k)$

Equating the x and y components gives: $\frac{1 + x_1}{2} = 0 \implies x_1 = -1$ $\frac{2 + y_1}{2} = k \implies y_1 = 2k - 2$

So, the point $R$ has coordinates $(-1, 2k - 2)$.

Since $R$ lies on the circle $x^2 + y^2 + x - 3y = 0$, we substitute $x = -1$ and $y = 2k - 2$ into the circle's equation: $(-1)^2 + (2k - 2)^2 + (-1) - 3(2k - 2) = 0$

Expanding and simplifying the equation: $1 + (4k^2 - 8k + 4) - 1 - 6k + 6 = 0$ $4k^2 - 14k + 10 = 0$ $2k^2 - 7k + 5 = 0$

Factoring the quadratic equation: $(2k - 5)(k - 1) = 0$

This gives two values for $k$: $k_1 = 1 \quad \text{and} \quad k_2 = \frac{5}{2}$

Now, we calculate the coordinates of the two endpoints $R$ and $S$:

1. For $k_1 = 1$: $y_1 = 2(1) - 2 = 0 \implies R(-1, 0)$

2. For $k_2 = \frac{5}{2}$: $y_2 = 2\left(\frac{5}{2}\right) - 2 = 3 \implies S(-1, 3)$

The midpoint $(\alpha, \beta)$ of the line segment $RS$ connecting $R(-1, 0)$ and $S(-1, 3)$ is: $\alpha = \frac{-1 + (-1)}{2} = -1$ $\beta = \frac{0 + 3}{2} = \frac{3}{2}$

Now, we evaluate $6(\alpha + \beta)$: $6(\alpha + \beta) = 6 \left( -1 + \frac{3}{2} \right) = 6 \left( \frac{1}{2} \right) = 3$

Thus, the required value of $6(\alpha + \beta)$ is $3$.