Find the equation of a straight line perpendicular to the line 3x - 4y + 12 = 0 and having same y-intercept as 2x - y + 5 = 0.

Given equation of line,

⇒ 3x - 4y + 12 = 0

Converting it in the form y = mx + c,

⇒ 4y = 3x + 12

⇒ y = $\dfrac{3}{4}x + 3$.

Comparing with y = mx + c we get,

Slope (m₁) = $\dfrac{3}{4}$.

Let the slope of the line perpendicular to the given line be m₂. So,

m₁ × m₂ = -1

$\Rightarrow \dfrac{3}{4} \times m$2 = -1 \\[1em] \Rightarrow m2 = -\dfrac{4}{3}.

The other line is 2x - y + 5 = 0 or y = 2x + 5.

Comparing with y= mx + c we get, c = 5.

So, the new line has slope = $-\dfrac{4}{3}$ and y-intercept = 5.

Putting these values in y = mx + c,

$\Rightarrow y = -\dfrac{4}{3}x + 5 \\[1em] \Rightarrow y = \dfrac{-4x + 15}{3} \\[1em] \Rightarrow 3y = -4x + 15 \\[1em] \Rightarrow 4x + 3y = 15 \\[1em] \Rightarrow 4x + 3y - 15 = 0.$

Hence, the equation of the line is 4x + 3y - 15 = 0.