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 $\mathrm{\alpha }$ with the positive x-axis.
Then, we have,
$\mathrm{tan\alpha }=\frac{5}{12}$

Here, $\tan \left({180}^{\circ }+\alpha \right)=\mathrm{tan\alpha }$

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




$$\begin{gathered} \mathrm{Now},\mathrm{tan\alpha }=\frac{5}{12} \\ \Rightarrow \mathrm{sin\alpha }=\frac{5}{13}\mathrm{and}\mathrm{cos\alpha }=\frac{12}{13} \end{gathered}$$

Here, p = 2

So, the equations of the lines in normal form are

$$\begin{gathered} x\mathrm{cos\alpha }+y\mathrm{sin\alpha }=p\mathrm{and}x\cos \left({180}^{\circ }+\alpha \right)+y\sin \left({180}^{\circ }+\mathrm{\alpha }\right)=p \\ \Rightarrow x\mathrm{cos\alpha }+y\mathrm{sin\alpha }=2\mathrm{and}-x\mathrm{cos\alpha }-y\mathrm{sin\alpha }=2 \\ \Rightarrow \frac{12x}{13}+\frac{5y}{13}=2\mathrm{and}-\frac{12x}{13}-\frac{5y}{13}=2 \\ \Rightarrow 12x+5y=26\mathrm{and}12x+5y=-26 \end{gathered}$$