Find the equation of a line whose

(i) slope = 3, y-intercept = -5.

(ii) slope = $-\dfrac{2}{7}$, y-intercept = 3.

(iii) gradient = $\sqrt{3}$, y-intercept = $-\dfrac{4}{3}$

(iv) inclination = 30°, y-intercept = 2.

(i) The equation of the straight line is given by,

y = mx + c, where m is the slope and c is the y-intercept.

Given slope = 3 and y-intercept = -5. Putting values in equation we get,

y = 3x - 5.

Hence, the equation of the straight line is y = 3x - 5.

(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 = $-\dfrac{2}{7}$ and y-intercept = 3. Putting values in equation we get,

$\Rightarrow y = -\dfrac{2}{7}x + 3 \\[1em] \Rightarrow y = \dfrac{-2x + 21}{7} \\[1em] \Rightarrow 7y = -2x + 21 \\[1em] \Rightarrow 2x + 7y - 21 = 0.$

Hence, the equation of straight line is 2x + 7y - 21 = 0.

(iii) 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 = $-\dfrac{4}{3}$. Putting values in equation we get,

$\Rightarrow y = \sqrt{3}x + \Big(-\dfrac{4}{3}\Big) \\[1em] \Rightarrow y = \dfrac{3\sqrt{3}x - 4}{3} \\[1em] \Rightarrow 3y = 3\sqrt{3}x - 4 \\[1em] \Rightarrow 3\sqrt{3}x - 3y - 4 = 0.$

Hence, the equation of straight line is $3\sqrt{3}x - 3y - 4 = 0.$

(iv) The equation of straight line is given by,

y = mx + c, where m is the slope and c is the y-intercept.

Given inclination = θ = 30° and y-intercept = 2.
Slope = m = tan θ = tan 30° = $\dfrac{1}{\sqrt{3}}$

Putting values in equation we get,

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

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