Find the equation of a line, which has the y-intercept 4, and is parallel to the line 2x - 3y - 7 = 0. Find the coordinates of the point where it cuts the x-axis.

Given equation of line,

⇒ 2x - 3y - 7 = 0,

Converting it in the form y = mx + c,

⇒ 3y = 2x - 7

⇒ y = $\dfrac{2}{3}x - \dfrac{7}{3}$.

Comparing with y = mx + c, slope = $\dfrac{2}{3}$.

Since,the other line is parallel so, its slope will also be equal to $\dfrac{2}{3}$. Given, y-intercept is 4 or c = 4.

Putting values of slope and y-intercept in y = mx + c, we will get the equation of line as,

⇒ y = $\dfrac{2}{3}x + 4$

⇒ y = $\dfrac{2x + 12}{3}$

⇒ 3y = 2x + 12

⇒ 2x - 3y + 12 = 0.

At the point where the line intersects the x-axis, the y-coordinate there will be zero. So, putting y = 0 in 2x - 3y + 12 = 0.

⇒ 2x - 3(0) + 12 = 0
⇒ 2x = -12
⇒ x = -6.

∴ Coordinates = (-6, 0).

Hence, the equation of the line is 2x - 3y + 12 = 0 and it intersects the x-axis at (-6, 0).