Give line is 5x−12y+6=0 Slope of this line is m1=‌
−5
−12
=‌
5
12
Since, line L is perpendicular to this line, the product of their slope is -1 . Let the slope of line L be mL. Then, mL×‌
5
12
=−1 ⇒‌‌mL=−‌
12
5
The equation of a line in normal form is x‌cos‌θ+ysin‌θ=P, where P is the distance from the origin to the line and θ is the angle made by the perpendicular from the origin. We have P=2 unit, So, the equation of line is x‌cos‌θ+ysin‌θ=2 Slope, mL=‌
−cos‌θ
sin‌θ
=−cot‌θ ∴−cot‌θ=‌
−12
5
⇒cot‌θ=‌
12
5
For y-intercept, x=0 then ysin‌θ=2 ⇒‌‌y=‌
2
sin‌θ
And since y-intercept is positive, so ‌‌
2
sin‌θ
>0 ‌⇒sin‌θ>0 ⇒cot‌θ=‌
12
5
>0,θ must be in first quadrant. Now, tan‌θ=‌
