A line, with the slope greater than one, passes through the point $$A(4, 3)$$ and intersects the line $$x - y - 2 = 0$$ at the point $$B$$. If the length of the line segment $$AB$$ is $$\dfrac{\sqrt{29}}{3}$$, then $$B$$ also lies on the line

A line with slope greater than 1 passes through $$ A(4, 3) $$ and intersects $$ x - y - 2 = 0 $$ at point $$ B $$. The length $$ AB = \frac{\sqrt{29}}{3} $$. We need to find which line $$ B $$ also lies on.

Let the slope of the line be $$ m $$ where $$ m > 1 $$. The line through $$ A(4, 3) $$ is:

$$y - 3 = m(x - 4)$$

On the line $$ x - y - 2 = 0 $$, we have $$ y = x - 2 $$. Substituting:

$$x - 2 - 3 = m(x - 4)$$

$$x - 5 = mx - 4m$$

$$x(1 - m) = 5 - 4m$$

$$x = \frac{5 - 4m}{1 - m} = \frac{4m - 5}{m - 1}$$

$$y = x - 2 = \frac{4m - 5}{m - 1} - 2 = \frac{4m - 5 - 2m + 2}{m - 1} = \frac{2m - 3}{m - 1}$$

$$AB^2 = (x_B - 4)^2 + (y_B - 3)^2 = \frac{29}{9}$$

$$x_B - 4 = \frac{4m - 5}{m - 1} - 4 = \frac{4m - 5 - 4m + 4}{m - 1} = \frac{-1}{m - 1}$$

$$y_B - 3 = \frac{2m - 3}{m - 1} - 3 = \frac{2m - 3 - 3m + 3}{m - 1} = \frac{-m}{m - 1}$$

$$AB^2 = \frac{1}{(m-1)^2} + \frac{m^2}{(m-1)^2} = \frac{1 + m^2}{(m-1)^2} = \frac{29}{9}$$

$$9(1 + m^2) = 29(m - 1)^2$$

$$9 + 9m^2 = 29m^2 - 58m + 29$$

$$20m^2 - 58m + 20 = 0$$

$$10m^2 - 29m + 10 = 0$$

Using the quadratic formula:

$$m = \frac{29 \pm \sqrt{841 - 400}}{20} = \frac{29 \pm \sqrt{441}}{20} = \frac{29 \pm 21}{20}$$

$$m = \frac{50}{20} = \frac{5}{2} \quad \text{or} \quad m = \frac{8}{20} = \frac{2}{5}$$

Since slope $$ m > 1 $$, we take $$ m = \frac{5}{2} $$.

$$x_B = \frac{4 \cdot \frac{5}{2} - 5}{\frac{5}{2} - 1} = \frac{10 - 5}{\frac{3}{2}} = \frac{5}{\frac{3}{2}} = \frac{10}{3}$$

$$y_B = x_B - 2 = \frac{10}{3} - 2 = \frac{4}{3}$$

So $$ B = \left(\frac{10}{3}, \frac{4}{3}\right) $$.

Option A: $$ 2x + y = 9 $$: $$ 2 \cdot \frac{10}{3} + \frac{4}{3} = \frac{24}{3} = 8 \neq 9 $$ ✗

Option B: $$ 3x - 2y = 7 $$: $$ 3 \cdot \frac{10}{3} - 2 \cdot \frac{4}{3} = 10 - \frac{8}{3} = \frac{22}{3} \neq 7 $$ ✗

Option C: $$ x + 2y = 6 $$: $$ \frac{10}{3} + 2 \cdot \frac{4}{3} = \frac{10}{3} + \frac{8}{3} = \frac{18}{3} = 6 $$ ✓

Point $$ B $$ lies on the line $$ x + 2y = 6 $$, which corresponds to Option C.