Assuming:
The perpendicular drawn from the origin make acute angle α with the positive x–axis. Then, we have, tanα = \(\frac{5}{12}\)
We know that, tan(180∘ + α) = tanα
So, there are two possible lines, AB and CD, on which the perpendicular drawn from the origin has a slope equal to 5/12 .
Given:
Now tan α = \(\frac{5}{12}\)
⇒ sin α = \(\frac{5}{13}\) and cos α = \(\frac{12}{13}\)
Explanation:
So, the equations of the lines in normal form are
Formula Used: x cos α + y sin α = p
⇒ x cos α + y sin α = p and x cos(180° + α) + ysin(180° + α) = p
⇒ x cos α + y sin α = 2 and –x cos α – ysin α = 2 cos (180° + θ) = – cos θ , sin (180° + θ) = – sin θ
⇒ \(\frac{12x}{13}+\frac{5y}{13}\)
and 12x + 5y = – 26
Hence, the equation of line in normal form is \(\frac{12x}{13}+\frac{5y}{13}\) = 26 and 12x + 5y = – 26
