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 need to find the intersection points \(A\) and \(B\), determine the equation of the obtuse angle bisector of \(L_2\) and \(L_3\), find its intersection \(C\) with \(L_1\), and then compute the required ratio of distances.
Step 2: Key Formula or Approach:
For lines \(a_1 x + b_1 y + c_1 = 0\) and \(a_2 x + b_2 y + c_2 = 0\) with \(c_1, c_2 \ge 0\), the angle bisectors are given by \(\frac{a_1 x + b_1 y + c_1}{\sqrt{a_1^2 + b_1^2}} = \pm \frac{a_2 x + b_2 y + c_2}{\sqrt{a_2^2 + b_2^2}}\).
If \(a_1 a_2 + b_1 b_2>0\), the '+' sign corresponds to the obtuse angle bisector.
Step 3: Detailed Explanation:
First, find points \(A\) and \(B\):
Intersection of \(L_1: x = -3\) and \(L_2: x - y = 0 \implies y = -3\). So, \(A(-3, -3)\).
Intersection of \(L_1: x = -3\) and \(L_3: 3x + y = 0 \implies y = 9\). So, \(B(-3, 9)\).
Now, let's find the angle bisectors of \(L_2\) and \(L_3\).
\(L_2: x - y = 0\) and \(L_3: 3x + y = 0\).
We check the condition \(a_1 a_2 + b_1 b_2 = (1)(3) + (-1)(1) = 3 - 1 = 2>0\).
Since the value is positive, the obtuse angle bisector is given by the '+' sign:
\[ \frac{x - y}{\sqrt{1^2 + (-1)^2}} = +\frac{3x + y}{\sqrt{3^2 + 1^2}} \] \[ \frac{x - y}{\sqrt{2}} = \frac{3x + y}{\sqrt{10}} \implies \sqrt{5}(x - y) = 3x + y \] \[ (\sqrt{5} - 3)x = (\sqrt{5} + 1)y \implies y = \frac{\sqrt{5} - 3}{\sqrt{5} + 1} x \] Let's find point \(C\), the intersection of this bisector with \(L_1: x = -3\):
\[ y_C = \frac{\sqrt{5} - 3}{\sqrt{5} + 1} (-3) = \frac{9 - 3\sqrt{5}}{\sqrt{5} + 1} \times \frac{\sqrt{5} - 1}{\sqrt{5} - 1} = \frac{9\sqrt{5} - 9 - 15 + 3\sqrt{5}}{4} \] \[ y_C = \frac{12\sqrt{5} - 24}{4} = 3\sqrt{5} - 6 \] So, \(C(-3, 3\sqrt{5} - 6)\).
Now, we calculate the lengths \(AC\) and \(BC\):
Since all points \(A\), \(B\), and \(C\) lie on the vertical line \(x = -3\), the distances are simply the differences in their y-coordinates.
\[ AC = |(3\sqrt{5} - 6) - (-3)| = |3\sqrt{5} - 3| = 3(\sqrt{5} - 1) \] \[ BC = |9 - (3\sqrt{5} - 6)| = |15 - 3\sqrt{5}| = 3(5 - \sqrt{5}) = 3\sqrt{5}(\sqrt{5} - 1) \] The ratio \(BC : AC\) is:
\[ \frac{BC}{AC} = \frac{3\sqrt{5}(\sqrt{5} - 1)}{3(\sqrt{5} - 1)} = \sqrt{5} \] Step 4: Final Answer:
The required square ratio \(BC^2 : AC^2\) is:
\[ (\sqrt{5})^2 : 1^2 = 5 : 1 \]