$\begin {array} {1 1} (A)\;\large\frac{1}{3} & \quad (B)\;\large\frac{2}{3} \\ (C)\;1 & \quad (D)\;\large\frac{4}{3} \end {array}$

- Slope of a line which is perpendicular to a line having slope 'm' is $ - \large\frac{1}{m}$.
- Equation of a line having slope $m$ and passing through $(x_1, y_1)$ is $ (y-y_1)=m(x-x_1)$

Step 1 :

Equation of the given line is $ - \large\frac{3}{1}$$ = -3$

Hence slope of the line which is perpendicular to the above line is $ \large\frac{1}{3}$

It is given that the line passes through the point (2,2)

Hence equation of the required line is $(y-2)=\large\frac{1}{3}$$(x-2)$

$ \Rightarrow 3y-6=x-2$

(i.e) $3y=x+4$

$ \therefore y = \large\frac{1}{3}$$x+ \large\frac{4}{3}$

Hence the y intercept is $ \large\frac{4}{3}$

Hence 'D' is the correct option.

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