Question

Let the line \(L_1 : x + 3 = 0\) intersect the lines \(L_2 : x - y = 0\) and \(L_3 : 3x + y = 0\) at the points A and B, respectively. Let the bisector of the obtuse angle between the lines \(L_2\) and \(L_3\) intersect the line \(L_1\) at the point C. Then \(BC^2 : AC^2\) is equal to:

• A. 5:1
• B. 1:5
• C. 2:3
• D. 3:2

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We first find the coordinates of points $A$ and $B$ by solving the intersections. Then, we determine the equation of the angle bisector between $L_2$ and $L_3$ and find its intersection with $L_1$ to get point $C$. Finally, we calculate the distances.

Step 2: Key Formula or Approach:
Angle Bisector formula for $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$: \[ \frac{a_1x + b_1y + c_1}{\sqrt{a_1^2 + b_1^2}} = \pm \frac{a_2x + b_2y + c_2}{\sqrt{a_2^2 + b_2^2}} \]

Step 3: Detailed Explanation:
1. Point A: $x = -3$ and $x - y = 0 \implies y = -3$. So $A(-3, -3)$. 2. Point B: $x = -3$ and $3x + y = 0 \implies y = 9$. So $B(-3, 9)$. 3. Angle Bisector: $L_2: x - y = 0$ and $L_3: 3x + y = 0$. \[ \frac{x-y}{\sqrt{2}} = \pm \frac{3x+y}{\sqrt{10}} \implies \sqrt{5}(x-y) = \pm(3x+y) \] 4. For the obtuse bisector, we check the sign of $a_1a_2 + b_1b_2 = (1)(3) + (-1)(1) = 2>0$. The (+) sign gives the obtuse bisector: \[ \sqrt{5}x - \sqrt{5}y = 3x + y \implies (\sqrt{5}-3)x = (1+\sqrt{5})y \] 5. Point C: Intersect with $x = -3$: \[ (\sqrt{5}-3)(-3) = (1+\sqrt{5})y \implies y = \frac{-3\sqrt{5}+9}{1+\sqrt{5}} = 3\sqrt{5}-6 \] 6. Calculate $BC^2$ and $AC^2$ (since all have same $x$, we only compare $y$-differences): \[ AC^2 = (y_C - (-3))^2 = (3\sqrt{5}-3)^2 = 9(5+1-2\sqrt{5}) = 9(6-2\sqrt{5}) \] \[ BC^2 = (y_C - 9)^2 = (3\sqrt{5}-15)^2 = 9(5+25-10\sqrt{5}) = 9(30-10\sqrt{5}) \] 7. Ratio $BC^2 / AC^2 = \frac{30-10\sqrt{5}}{6-2\sqrt{5}} = \frac{5(6-2\sqrt{5})}{6-2\sqrt{5}} = 5$.

Step 4: Final Answer:
The ratio $BC^2 : AC^2$ is 5:1.