Lines L1 : y - x = 0 and L2 : 2x + y = 0 intersect the line L3 : y + 2 = 0 at P and Q, respectively.

Q: Wait, why can't I just find the equation of the angle bisector and solve for $R$ directly? Isn't that safer?

You absolutely can! That is the 'brute force' method, and it will give you the correct answer. However, in a high-pressure exam like JEE Advanced, time is your most precious resource. The Angle Bisector Theorem is a powerful tool that bypasses the need for finding the line equation of the bisector entirely. Always look for the geometric shortcut before diving into heavy algebra.

Q: How do I know if I should use the internal or external angle bisector theorem?

The problem explicitly mentions the 'acute angle' between the lines. In the context of the triangle $\triangle APQ$, the bisector of the angle at $A$ that hits the segment $PQ$ is the internal bisector. The external bisector would intersect the extension of the line $PQ$, which is not what we are dealing with here. Always sketch the figure to confirm!

Q: Why is Statement 2 false? Doesn't an angle bisector always create similar triangles?

That is a common misconception! An angle bisector divides the opposite side in a specific ratio, but it does not guarantee that the two resulting triangles are similar. For $\triangle APR$ to be similar to $\triangle AQR$, their corresponding angles must be equal. As we saw, the base angles of $\triangle APQ$ are different, so the bisector cannot create similar triangles. Similarity is a much stronger condition than just having a bisector.

Question

Lines $L_1 : y - x = 0$ and $L_2 : 2x + y = 0$ intersect the line $L_3 : y + 2 = 0$ at $P$ and $Q$, respectively. The bisector of the acute angle between $L_1$ and $L_2$ intersects $L_3$ at $R$.
STATEMENT-1 : The ratio $PR : RQ$ equals $2\sqrt{2} : \sqrt{5}$.
because
STATEMENT-2 : In any triangle, bisector of an angle divides the triangle into two similar triangles.

Options:

(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

(C) Statement-1 is True, Statement-2 is False

(D) Statement-1 is False, Statement-2 is True

Sigma Percentile · Accepted Answer

### The Geometry of the Playground

Welcome, future engineer. Today, we are not just solving a coordinate geometry problem; we are exploring the elegant interplay between lines and triangles.

Imagine you are standing on a coordinate plane with three lines: 
$L_1: y - x = 0$, 
$L_2: 2x + y = 0$, 
and $L_3: y + 2 = 0$.

These lines define the boundaries of a playground. $L_1$ is a line passing through the origin with a slope of $1$, $L_2$ is a steeper line with a slope of $-2$, and $L_3$ is a horizontal floor at $y = -2$.

### The Intersection Hunt

First, we identify the vertices of the triangle. The lines $L_1$ and $L_2$ intersect at the origin, $A(0,0)$.

Next, we find point $P$, where $L_1$ intersects the floor $L_3$. Since $L_1$ is $y = x$ and $L_3$ is $y = -2$, we find $P(-2, -2)$.

Similarly, we find point $Q$, where $L_2$ intersects the floor $L_3$. Substituting $y = -2$ into $2x + y = 0$, we get $2x - 2 = 0$, which yields $x = 1$. Thus, $Q(1, -2)$.

We have successfully defined our triangle $	riangle APQ$ with vertices $A(0,0)$, $P(-2,-2)$, and $Q(1,-2)$.

### The Power of the Angle Bisector Theorem

The problem introduces the bisector of the acute angle at $A$, which intersects $L_3$ at point $R$. While one could derive the equation of this bisector, we utilize the **Angle Bisector Theorem** for efficiency.

The theorem states that the internal bisector of an angle of a triangle divides the opposite side in the ratio of the sides containing the angle. Mathematically, this is expressed as:

$$\frac{PR}{RQ} = \frac{AP}{AQ}$$

### Calculating the Sides

We calculate the lengths $AP$ and $AQ$ using the distance formula from the origin $A(0,0)$:

$$AP = \sqrt{(-2 - 0)^2 + (-2 - 0)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$$

$$AQ = \sqrt{(1 - 0)^2 + (-2 - 0)^2} = \sqrt{1 + 4} = \sqrt{5}$$

With these lengths, the ratio $\frac{PR}{RQ}$ is simply $\frac{2\sqrt{2}}{\sqrt{5}}$. Therefore, Statement 1 is **true**.

### Debunking the Similarity Myth

Finally, we evaluate Statement 2, which claims that an angle bisector always divides a triangle into two similar triangles.

For $	riangle APR$ to be similar to $	riangle AQR$, their corresponding angles must be equal. Specifically, $\angle APR$ must equal $\angle AQR$.

However, $\angle APR$ is the angle $L_1$ makes with the horizontal, which is $45^\circ$. Meanwhile, $\angle AQR$ is determined by the slope of $L_2$ (which is $-2$), resulting in an angle of $	an^{-1}(2) \approx 63.4^\circ$.

Since $45^\circ 
eq 63.4^\circ$, the triangles are not similar. Consequently, Statement 2 is **false**.

Select Answer:

Visualized Solution

Visualizing the Given Lines

Finding the Intersection Point $A$

Finding the Intersection Point $P$

Coordinates of Point $P$

Finding the Intersection Point $Q$

Coordinates of Point $Q$

The Triangle $A P Q$

Calculating Length of $A P$

Calculating Length of $A Q$

The Angle Bisector $A R$

Internal Angle Bisector Theorem

Evaluating Statement-1

Evaluating Statement-2

Conclusion

The Sigma Insight: Angle Between Two Lines

The Geometry of the Playground

The Intersection Hunt

The Power of the Angle Bisector Theorem

Calculating the Sides

Debunking the Similarity Myth

Similar Questions

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Select Answer:

Visualized Solution

Visualizing the Given Lines

Finding the Intersection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Finding the Intersection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Coordinates of Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Finding the Intersection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Coordinates of Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

The Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } APQ

Calculating Length of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AP

Calculating Length of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AQ

The Angle Bisector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AR