For the line \( 3x + 4y - 12 = 0 \), rewrite in slope-intercept form \( y = mx + c \): \[ 4y = -3x + 12 \implies y = -\frac{3}{4}x + 3 \] The slope of the given line is \( m = -\frac{3}{4} \).
The slope of a line perpendicular to it is the negative reciprocal: \[ m_{\text{perpendicular}} = -\frac{1}{-\frac{3}{4}} = \frac{4}{3} \] Thus, the slope is: \[ \boxed{\frac{4}{3}} \]
| List-I | List-II | ||
|---|---|---|---|
| (A) | $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ | (I) | $[\frac{1}{3} , 1 ]$ |
| (B) | $(x)=|[x]|,x \in [R$ | (II) | Z |
| (C) | $h(x) = |x - [x]| , x \in [R$ | (III) | W |
| (D) | $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ | (IV) | [0, 1) |
| (V) | { -1, 1} | ||
| List I | List II | ||
|---|---|---|---|
| (A) | $\lambda=8, \mu \neq 15$ | 1. | Infinitely many solutions |
| (B) | $\lambda \neq 8, \mu \in R$ | 2. | No solution |
| (C) | $\lambda=8, \mu=15$ | 3. | Unique solution |