Question

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 1

Let the perpendicular drawn from the origin make acute angle \[\alpha\] with the positive x-axis.
Then, we have,

\[\text { tan }\alpha = \frac{5}{12}\]

Here, 

\[\tan\left( {180}^\circ + \alpha \right) = \text { tan }\alpha\]

So, there are two possible lines, AB and CD, on which the perpendicular drawn from the origin has slope equal to \[\frac{5}{12}\].

\[\text { Now }, tan\alpha = \frac{5}{12}\]

\[ \Rightarrow \text { sin }\alpha = \frac{5}{13} \text { and cos }\alpha = \frac{12}{13}\]

Here, p = 2
So, the equations of the lines in normal form are

\[x\text {cos }\alpha + y\text { sin }\alpha = p \text { and } x\cos\left( {180}^\circ + \alpha \right) + y\text { sin }\left( {180}^\circ + \alpha \right) = p\]

\[ \Rightarrow x\text { cos }\alpha + y\text { sin }\alpha = 2 \text { and} - \text { x cos }\alpha - ysin\alpha = 2\]

\[ \Rightarrow \frac{12x}{13} + \frac{5y}{13} = 2 \text { and } - \frac{12x}{13} - \frac{5y}{13} = 2\]

\[ \Rightarrow 12x + 5y = 26 \text { and } 12x + 5y = - 26\]