Find the equation of a line whose:

(i) Slope = $\dfrac{3}{4}$ and y-intercept = –4

(ii) Gradient = $\sqrt{3}$ and y-intercept = $\dfrac{-2}{3}$

(i) The equation of the straight line is given by, y = mx + c, we get where m is the slope and c is the y-intercept.

Given slope = $\dfrac{3}{4}$ and y-intercept = -4. Substituting values in equation we get,

⇒ y = $\dfrac{3}{4}$x - 4.

⇒ y = $\dfrac{3x - 16}{4}$

⇒ 4y = 3x - 16

⇒ 3x - 4y = 16.

Hence, equation of the line is 3x - 4y = 16.

(ii) The equation of straight line is given by y = mx + c, where m is the slope and c is the y-intercept.

Given slope = $\sqrt{3}$ and y-intercept = $\dfrac{-2}{3}$.

Substituting values in equation we get,

$\Rightarrow y = \sqrt{3}x + \Big(\dfrac{-2}{3}\Big) \\[1em] \Rightarrow y = \dfrac{3\sqrt{3}x -2}{3} \\[1em] \Rightarrow 3y = 3\sqrt{3}x - 2 \\[1em] \Rightarrow 3\sqrt{3}x -3y - 2 = 0.$

Hence, equation of the line is $3\sqrt{3}x -3y - 2 = 0$.