Two straight lines 3x - 2y = 15 and 2x + ky + 8 = 0.

Assertion (A) : The given two lines are perpendicular to each other and k = 3.

Reason (R) : If the inclination of two lines are α and β; then tan α = -cot β.

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true and R is correct reason for A.

4. Both A and R are true and R is incorrect reason for A.

Given, two lines: 3x - 2y = 15 and 2x + ky + 8 = 0

Converting the equation of first line in slope-intercept form (y = mx + c), we get :

⇒ 3x - 2y = 15

⇒ -2y = -3x + 15

⇒ y = $\dfrac{3}{2}x - \dfrac{15}{2}$

∴ Slope of the line (m₁) = $\dfrac{3}{2}$

Converting the equation of second line in slope-intercept form (y = mx + c), we get :

⇒ 2x + ky + 8 = 0

⇒ ky = -2x - 8

⇒ y = $-\dfrac{2}{k}x - \dfrac{8}{k}$

∴ Slope of the line (m₂) = $-\dfrac{2}{k}$

We know that,

The two lines are perpendicular if product of their slopes is -1.

$\Rightarrow m$1 \times m2 = -1\\[1em] \Rightarrow \dfrac{3}{2} \times \Big(-\dfrac{2}{k}\Big) = -1\\[1em] \Rightarrow -\dfrac{3}{k} = -1\\[1em] \Rightarrow -3 = -k\\[1em] \Rightarrow k = 3\\[1em]

So, assertion (A) is true.

If the inclination of two lines are α and β, then Slope of first line (m₁) = tan α and Slope of second line (m₂) = tan β.

⇒ m₁ x m₂ = -1

⇒ tan α x tan β = -1

⇒ tan α = $-\dfrac{1}{\text{tan } β}$

⇒ tan α = -cot β

So, reason (R) is true, but it is not the correct reason for assertion (A).

Hence, option 4 is the correct option.