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

$\text{Let the perpendicular drawn from the origin make acute angle}\alpha \text{with the positive x-axis.}$

$\text{Then, we have}$

$tan\alpha =\frac{5}{12}$

$\text{Here,}tan({180}^{\circ }+\alpha )=tan\alpha$

$\text{So, there are two possible lines, AB and CD, on which the perpendicular drawn}$

$\text{from the origin has slope equal to}\frac{5}{12}$

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

$\Rightarrow sin\alpha =\frac{5}{13}andcos\alpha =\frac{12}{13}$

$\text{Here,}p=2$

$\text{So, the equations of the lines in normal form are}$

$xcos\alpha +ysin\alpha =pandxcos({180}^{\circ }+\alpha )+ysin({180}^{\circ }+\alpha )=p$

$\Rightarrow xcos\alpha +ysin\alpha =2and-xcos\alpha -ysin\alpha =2$

$\Rightarrow \frac{12x}{13}+\frac{5y}{13}=2and-\frac{12x}{13}-\frac{5y}{13}=2$

$\Rightarrow 12x+5y=26and12x+5y=-26$