Find the equation of the straight line upon which the length of the pe

1.	Find the equation of the straight line upon which the length of the perpendicular from the origin is 2, and the slope of this perpendicular is \(\frac{5}{12}\)
Answer» 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

Find the equation of the straight line upon which the length of the perpendicular from the origin is 2, and the slope of this perpendicular is \(\frac{5}{12}\)

Answer»

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