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Step 1: Identify and Write Down the Given Information
• We have a triangle $ \triangle ABC $.
• The angle bisector of $ \angle B $ is given by the line $ y = x $.
• The side $ AC $ is given by $ 2x - y = 2 $.
• The coordinates of $ A $ are $ (4, 6) $, and the coordinates of $ B $ are $ (\alpha, \beta) $.
• It is given that $ 2 \, AB = BC $.
• We wish to find the value of $ \alpha + 2\beta $.
Step 2: Find the Intersection of the Angle Bisector and Side AC
The line of the angle bisector is $ y = x $, and the equation for $ AC $ is $ 2x - y = 2 $.
Substitute $y = x$ into $2x - y = 2$:
$ 2x - x = 2 \quad \Longrightarrow \quad x = 2. $
Then $ y = x = 2 $.
Therefore, the point of intersection (call it $D$) is $ D(2, 2) $.
Step 3: Use the Angle Bisector Theorem
The angle bisector theorem states that the bisector of an angle in a triangle divides the opposite side in the ratio of the adjacent sides. Here:
$ \frac{AD}{CD} = \frac{AB}{BC}. $
We are also given: $ 2 \, AB = BC \quad \Longrightarrow \quad \frac{AB}{BC} = \frac{1}{2}. $
Thus,
$ \frac{AD}{CD} = \frac{1}{2} \quad \Longrightarrow \quad AD : CD = 1 : 2. $
Step 4: Determine the Coordinates of C Using the Section Formula
Let $ C(h,k) $. Since $D$ divides $AC$ in the ratio $1 : 2$ (with $A$ corresponding to the portion $1$ and $C$ to the portion $2$), the coordinates of $D$ can be found using the section formula:
$ D \biggl(\dfrac{1 \cdot h + 2 \cdot 4}{1 + 2}, \, \dfrac{1 \cdot k + 2 \cdot 6}{1 + 2}\biggr) = (2, 2). $
Hence,
$ \dfrac{h + 8}{3} = 2, \quad \dfrac{k + 12}{3} = 2. $
Solving these:
$ h + 8 = 6 \quad \Longrightarrow \quad h = -2, $
$ k + 12 = 6 \quad \Longrightarrow \quad k = -6. $
Therefore, $ C(-2, -6) $.
Step 5: Express Distances AB and BC in Terms of $ \alpha $ and $ \beta $
We have:
$ A(4, 6) $ and $ B(\alpha, \beta) $, so
$ AB = \sqrt{(\alpha - 4)^2 + (\beta - 6)^2}. $
$ C(-2, -6) $ and $ B(\alpha, \beta) $, so
$ BC = \sqrt{(\alpha + 2)^2 + (\beta + 6)^2}. $
Given $ 2 \, AB = BC $, we can write:
$ \sqrt{(\alpha + 2)^2 + (\beta + 6)^2} = 2 \, \sqrt{(\alpha - 4)^2 + (\beta - 6)^2}. $
Square both sides to eliminate the square roots:
$ (\alpha + 2)^2 + (\beta + 6)^2 \;=\; 4 \bigl[(\alpha - 4)^2 + (\beta - 6)^2 \bigr]. $
Step 6: Solve for $ \alpha $ and $ \beta $
Expand and simplify. After ensuring consistency with the given geometric conditions (in particular, that $B$ is different from the intersection point $D(2,2)$), one finds a valid solution:
$ \alpha = 14, \quad \beta = 14. $
Step 7: Compute $ \alpha + 2\beta $
Substitute $ \alpha = 14 $ and $ \beta = 14 $:
$ \alpha + 2\beta = 14 + 2 \times 14 = 14 + 28 = 42. $
Hence, the required value is $ \boxed{42} $.
