Let the line intersect the lines and at the points A and B, respectively. Let the bisector of the obtuse angle between the lines and intersect the line at the point C. Then is equal to: [2026]

Question

Let the line $L_{1} : x + 3 = 0$ intersect the lines $L_{2} : x - y = 0$ and $L_{3} : 3 x + y = 0$ 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 $B C^{2} : A C^{2}$ is equal to: [2026]

Smart Achievers · Accepted Answer

(1)

We have, A : (–3, –3) and B : (–3, 9)

The equations of the angle bisectors given by:

$\frac{x - y}{\sqrt{1^{2} + {(- 1)}^{2}}} = \pm \frac{3 x + y}{\sqrt{3^{2} + 1^{2}}}$

$\Rightarrow \frac{x - y}{\sqrt{2}} = \pm \frac{3 x + y}{\sqrt{10}}$

$\Rightarrow \sqrt{5} (x - y) = \pm (3 x + y)$

To find the obtuse angle bisector, we check the sign of $a_{1} a_{2} + b_{1} b_{2}$ for the lines $L_{2} (x - y = 0)$ and $L_{3} (3 x + y = 0)$ , where $a_{1} = 1$ , $b_{1} = - 1$ , $a_{2} = 3$ , $b_{2} = 1$ : $a_{1} a_{2} + b_{1} b_{2} = (1) (3) + (- 1) (1) = 3 - 1 = 2$

Since, the result is positive, the obtuse angle bisector is given by the equation with the minus sign (opposite the sign of $a_{1} a_{2} + b_{1} b_{2}$ ):

$\sqrt{5} (x - y) = - (3 x + y)$

$\Rightarrow \sqrt{5} x - \sqrt{5} y = - 3 x - y$

$\Rightarrow y (1 - \sqrt{5}) = - x (3 x + \sqrt{5})$

$\Rightarrow y = x \frac{3 + \sqrt{5}}{\sqrt{5} - 1}$

Point C lies on $L_{1} = (x = - 3)$ , so substituting x = –3 into the bisector equations:

$y_{C} = (- 3) \frac{3 + \sqrt{5}}{\sqrt{5} - 1} = - 3 \frac{(3 + \sqrt{5}) (\sqrt{5} + 1)}{(\sqrt{5} - 1) (\sqrt{5} + 1)}$

$\Rightarrow y_{C} = - 3 \frac{3 \sqrt{5} + 3 + 3 + \sqrt{5}}{5 - 1}$

$= - 3 \frac{8 + 4 \sqrt{5}}{4} = - 3 (2 + \sqrt{5})$

$C = (- 3, - 6 - 3 \sqrt{5})$

$∴ \frac{B C^{2}}{A C^{2}} = \frac{90 (3 + \sqrt{5})}{18 (3 + \sqrt{5})} = 5 : 1$

Solution

1 5 : 1

2 1 : 5

3 2 : 3

4 3 : 2

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