Q.Two lines passing through the point $(2, 3)$ intersects each other at an angle of $60^\circ$. If slope of one line is $2$, find equation of the other line.

Concept: Slope Calculation — When two lines intersect at an angle $\theta$, the relation between their slopes $m_1$ and $m_2$ is given by:

Step 1: Here $\theta = 60^\circ$, so $\tan 60^\circ = \sqrt{3}$. Given $m_1 = 2$, let the other slope be $m$.

Step 2: Substitute into the formula:

Step 3: Remove the absolute value and solve both cases:

• Case 1: $\frac{m - 2}{1 + 2m} = \sqrt{3} \;\Rightarrow\; m - 2 = \sqrt{3}(1 + 2m)$ $\Rightarrow m - 2 = \sqrt{3} + 2\sqrt{3}m \;\Rightarrow\; m - 2\sqrt{3}m = \sqrt{3} + 2$ $\Rightarrow m(1 - 2\sqrt{3}) = \sqrt{3} + 2 \;\Rightarrow\; m = \frac{\sqrt{3} + 2}{1 - 2\sqrt{3}}$
• Case 2: $\frac{m - 2}{1 + 2m} = -\sqrt{3} \;\Rightarrow\; m - 2 = -\sqrt{3}(1 + 2m)$ …

When two lines intersect, the angle between them is determined by their slopes. The formula connecting the angle $\theta$ and the slopes $m_1$, $m_2$ is:

The absolute value is there because the angle between lines is always taken as the acute angle (or the smaller one, between $0^\circ$ and $90^\circ$). Here, $\theta = 60^\circ$, so $\tan 60^\circ = \sqrt{3}$.

We are given that one line has slope $m_1 = 2$ and passes through $(2,3)$. The other line also passes through $(2,3)$ — this is the intersection point. So once we find the slope $m_2$ of the second line, we can write its equation directly using the point-slope form.

Step-by-step solution

1. Set up the angle condition Using the formula:

2. Remove the absolute value — two cases The equation $\left| X \right| = \sqrt{3}$ means $X = \sqrt{3}$ or $X = -\sqrt{3}$. So:

3. Solve the first case

Bring $m_2$ terms together:

4. Solve the second case

Bring $m_2$ terms together:

5. Rationalise the denominators (optional but neat) For the first slope:

Numerator: $2(1) + 2(-2\sqrt{3}) - \sqrt{3}(1) - \sqrt{3}(-2\sqrt{3}) = 2 - 4\sqrt{3} - \sqrt{3} + 2\cdot 3 = 2 - 5\sqrt{3} + 6 = 8 - 5\sqrt{3}$

Denominator: $-11$

So $m_2 = \frac{8 - 5\sqrt{3}}{-11} = \frac{5\sqrt{3} - 8}{11}$.

For the second slope:

$m_2 = \frac{2 + \sqrt{3}}{1 - 2\sqrt{3}} \times \frac{1 + 2\sqrt{3}}{1 + 2\sqrt{3}} = \frac{(2 + \sqrt{3})(1 + 2\sqrt{3})}{1 - 12}$ …