Q.$P_1$, $P_2$ are points on either of the two lines $y-\sqrt{3}\,|x|=2$ at a distance of 5 units from their point of intersection. Find the coordinates of the foot of perpendiculars drawn from $P_1$, $P_2$ on the bisector of the angle between the given lines.

Concept: Slope Calculation — The given equation $y - \sqrt{3}\,|x| = 2$ represents two lines depending on the sign of $x$.

Step 1: Identify the lines and their intersection.

For $x \ge 0$, $y - \sqrt{3}x = 2 \Rightarrow y = \sqrt{3}x + 2$.

For $x < 0$, $y + \sqrt{3}x = 2 \Rightarrow y = -\sqrt{3}x + 2$.

Both lines have slope $\pm \sqrt{3}$, so they are inclined at $60^\circ$ and $120^\circ$ to the positive $x$-axis. Their point of intersection is found by solving: at $x=0$, both give $y=2$. So intersection is $O(0,2)$.

Step 2: Find points $P_1, P_2$ at distance 5 from $O$.

On $y = \sqrt{3}x + 2$, a point at distance 5 from $O$ satisfies $x^2 + (\sqrt{3}x)^2 = 25 \Rightarrow 4x^2 = 25 \Rightarrow x = \pm \frac{5}{2}$.

Since this line is for $x \ge 0$, take $x = \frac{5}{2}$, giving $P_1\left(\frac{5}{2}, \frac{5\sqrt{3}}{2} + 2\right)$.

On $y = -\sqrt{3}x + 2$ (for $x < 0$), similarly $x = -\frac{5}{2}$, giving $P_2\left(-\frac{5}{2}, \frac{5\sqrt{3}}{2} + 2\right)$.

Step 3: Foot of perpendicular on the angle bisector. …

The equation $y - \sqrt{3}\,|x| = 2$ actually represents two lines, because of the absolute value:

• For $x \ge 0$: $y - \sqrt{3}\,x = 2$ → $y = \sqrt{3}\,x + 2$
• For $x \le 0$: $y - \sqrt{3}(-x) = 2$ → $y = -\sqrt{3}\,x + 2$

These are two lines symmetric about the $y$-axis, both with slope $\pm \sqrt{3}$, meeting at the same $y$-intercept $(0,2)$.

Why this approach works

The key idea: when two lines are symmetric about a line (here the $y$-axis), that line is the angle bisector. The foot of the perpendicular from any point on one line onto the bisector is simply the projection along the horizontal direction — because the bisector is vertical. So instead of heavy coordinate geometry, we can use simple right-triangle geometry.

Step-by-step solution

1. Find the point of intersection

Both lines pass through $(0,2)$. That’s the only intersection.

2. Understand the geometry

Each line makes an angle of $60^\circ$ with the $x$-axis (since $\tan 60^\circ = \sqrt{3}$).

The $y$-axis (the bisector) is at $0^\circ$? No — careful: the $y$-axis is vertical, making $90^\circ$ with the $x$-axis. The two lines are at $60^\circ$ and $120^\circ$ from the positive $x$-axis. Their angle bisector is the line at $90^\circ$, i.e., the $y$-axis.

3. Locate points $P_1$ and $P_2$

We need points on each line at distance 5 from $(0,2)$.

Along the line $y = \sqrt{3}\,x + 2$, a point at distance $d$ from $(0,2)$ has coordinates:

• Horizontal shift: $d \cos 60^\circ = d \cdot \frac{1}{2}$
• Vertical shift: $d \sin 60^\circ = d \cdot \frac{\sqrt{3}}{2}$

So for $d=5$:

• $P_1$ (on the right branch): $\left( \frac{5}{2}, \, 2 + \frac{5\sqrt{3}}{2} \right)$

Similarly, on the left branch ($y = -\sqrt{3}\,x + 2$), the direction angle is $120^\circ$:

• Horizontal shift: $5 \cos 120^\circ = 5 \cdot \left(-\frac{1}{2}\right) = -\frac{5}{2}$ …