The equation of a line is $3x - 4y + 7 = 0$. Find the:
(a) slope of the line.
(b) equation of a line perpendicular to the given line and having a y-intercept equals to 2.
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(a) Given,
$3x - 4y + 7 = 0$
Writing in the form $y = mx + c$,
$$\begin{aligned}\Rightarrow 4y = 3x + 7 \\[1em] \Rightarrow y = \frac{3}{4}x + \frac{7}{4}.\end{aligned}$$
Comparing with $y = mx + c$, slope $m = \frac{3}{4}$.
Hence, the slope of the line = $\mathbf{\frac{3}{4}}$.
(b) We know that,
The product of the slopes of two perpendicular lines is −1.
Slope of the required line = $-\frac{1}{m} = -\frac{1}{\frac{3}{4}} = -\frac{4}{3}$.
Given, y-intercept $c = 2$.
By slope-intercept form, $y = mx + c$,
$$\begin{aligned}\Rightarrow y = -\frac{4}{3}x + 2 \\[1em] \Rightarrow 3y = -4x + 6 \\[1em] \Rightarrow 4x + 3y - 6 = 0.\end{aligned}$$
Hence, the equation of the required line is $\bm{4x + 3y - 6 = 0}$.