A line which is parallel to $y=x$ is rotated about the point $(2,0)$ through 15 degrees anti-clockwise.

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Find the y intercept of the line passing through the point of intersection of new line with y=x and at right angles to the new line

Since $y=x$ doesn’t pass through(2,0) I can only think of the line being shifted about (2,2). If I am wrong, it’s here.

The new angle becomes 60 degree with +x axis.

The slope of line perndicular to the new line is $\frac{-1}{\sqrt 3}$

It passes through (2,2) then $$y-2=\frac{-1}{\sqrt 3}(x-2)$$ $$x+y\sqrt 3 -2\sqrt 3 -2=0$$ So the y intercept is $$y=\frac{2+2\sqrt 3}{\sqrt 3}$$

But the answer is $y=4+2\sqrt 3 $

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Since new line is from point (2,0) and inclined at angle of π/3 with +ve x axis. Line is y=$\sqrt{3}$(x-2).$$ $$ Intersect line y=x at point A(3+$\sqrt{3}$,3+$\sqrt{3}$). $$ $$ Line from point A perpendicular to new line will have slope of $-\frac{1}{\sqrt{3}}$ line is $$y-(3+\sqrt{3})=-\frac{1}{\sqrt{3}}(x-(3+\sqrt{3}))$$ For y intercept put x=0 then $ y=\sqrt{3}+1+(3+\sqrt{3})$ $$y=4+2\sqrt{3}$$