Equation of y-axis is x = 0
line is 2x + 3y = 4
For their intersection, put x = 0 in 2x + 3y = 4
⇒ \(y =\frac 43\)
Hence, intersection point is \((0, \frac 43)\).
Angle of required line with +ve direction of x-axis is 60°.
\(\therefore \) Slope of required line is m = tan60° = √3.
Let required line is y = mx + C
⇒ \(y = \sqrt3 x + C\) \((\because m = \sqrt 3)\)
⇒ \(\frac 43 = \sqrt 3 \times 0 + C\) (By putting x = 0 & y = \(\frac 43\))
⇒ \(C = \frac 43\)
\(\therefore\) Equation of required line is \(y = \sqrt 3 x + \frac 43\).