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 …

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 $B{C}^{2}:A{C}^{2}$ is equal to:

Accepted Answer

Given:

${L}_{1}:x+3=0\Rightarrow x=-3$

${L}_{2}:x-y=0\Rightarrow y=x$

${L}_{3}:3x+y=0\Rightarrow y=-3x$

We first find points $A$ and $B$.

1. Coordinates of $A$
Point $A$ is the intersection of ${L}_{1}$ and ${L}_{2}$.

Since $x=-3$ and $y=x$, we get

$y=-3$

So,

$A=(-3,-3)$

2. Coordinates of $B$
Point $B$ is the intersection of ${L}_{1}$ and ${L}_{3}$.

Since $x=-3$ and $y=-3x$, we get

$y=-3(-3)=9$

So,

$B=(-3,9)$

3. Equation of angle bisectors of ${L}_{2}$ and ${L}_{3}$
The lines are:

${L}_{2}:x-y=0$

${L}_{3}:3x+y=0$

Angle bisectors are given by

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

That is,

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

We need the bisector of the obtuse angle.

Let us find its slope.

Since both lines pass through the origin, we can use the formula for slope of angle bisectors.

The slopes of ${L}_{2}$ and ${L}_{3}$ are:

for ${L}_{2}$, ${m}_{1}=1$

for ${L}_{3}$, ${m}_{2}=-3$

The acute angle bisector lies between the directions of these lines, and the obtuse angle bisector is the other one.

Now let us find the bisector which intersects $x=-3$ between $y=-3$ and $y=9$ or outside, by solving directly.

Substitute $x=-3$ into angle bisector equations.

4. Find point $C$
From

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

put $x=-3$:

$\frac{-3-y}{\sqrt{2}}=\frac{-9+y}{\sqrt{10}}$

Cross-multiplying:

$\sqrt{10}(-3-y)=\sqrt{2}(-9+y)$

This gives one point on ${L}_{1}$.

From

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

put $x=-3$:

$\frac{-3-y}{\sqrt{2}}=-\frac{-9+y}{\sqrt{10}}$

Instead of solving both, it is easier to use the fact that any point on the angle bisector is equidistant from the two lines.

Let $C=(-3,k)$ on ${L}_{1}$.

Distance of $C$ from ${L}_{2}:x-y=0$ is

$\frac{|x-y|}{\sqrt{2}}=\frac{|-3-k|}{\sqrt{2}}=\frac{|k+3|}{\sqrt{2}}$

Distance of $C$ from ${L}_{3}:3x+y=0$ is

$\frac{|3x+y|}{\sqrt{10}}=\frac{|-9+k|}{\sqrt{10}}=\frac{|k-9|}{\sqrt{10}}$

So,

$\frac{|k+3|}{\sqrt{2}}=\frac{|k-9|}{\sqrt{10}}$

Squaring:

$\frac{(k+3{)}^{2}}{2}=\frac{(k-9{)}^{2}}{10}$

$5(k+3{)}^{2}=(k-9{)}^{2}$

Expand:

$5({k}^{2}+6k+9)={k}^{2}-18k+81$

$5{k}^{2}+30k+45={k}^{2}-18k+81$

$4{k}^{2}+48k-36=0$

${k}^{2}+12k-9=0$

$k=-6\pm 3\sqrt{5}$

So the two bisectors meet ${L}_{1}$ at points with $y=-6+3\sqrt{5}$ and $y=-6-3\sqrt{5}$.

Now $A=(-3,-3)$ and $B=(-3,9)$.

The obtuse angle is the larger angle between ${L}_{2}$ and ${L}_{3}$.

Its bisector lies inside that larger region, which is the bisector with slope closer to negative small value, and it cuts $x=-3$ at the point between $A$ and $B$.

Since

$-6+3\sqrt{5}\approx 0.708$

this lies between $-3$ and $9$.

Hence,

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

5. Find $A{C}^{2}$ and $B{C}^{2}$
Since all three points have the same $x$-coordinate, distances depend only on $y$-coordinates.

$AC$
$AC=|-6+3\sqrt{5}-(-3)|=|-3+3\sqrt{5}|=3(\sqrt{5}-1)$

So,

$A{C}^{2}=9(\sqrt{5}-1{)}^{2}$

$BC$
$BC=|9-(-6+3\sqrt{5})|=|15-3\sqrt{5}|=3(5-\sqrt{5})$

So,

$B{C}^{2}=9(5-\sqrt{5}{)}^{2}$

Therefore,

$B{C}^{2}:A{C}^{2}=(5-\sqrt{5}{)}^{2}:(\sqrt{5}-1{)}^{2}$

Now,

$(5-\sqrt{5}{)}^{2}=25+5-10\sqrt{5}=30-10\sqrt{5}=10(3-\sqrt{5})$

and

$(\sqrt{5}-1{)}^{2}=5+1-2\sqrt{5}=6-2\sqrt{5}=2(3-\sqrt{5})$

Hence,

$B{C}^{2}:A{C}^{2}=10(3-\sqrt{5}):2(3-\sqrt{5})=5:1$

So the required ratio is

$5:1$

Hence, the correct option is .

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