Find the equation f the straight line at a distance of 3 units from the origin such that the perpendicular from the origin to the line makes and angle `alpha` given by `t a nalpha=5/(12)` with the positive direction of x-axis.

Explanation

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

`\implies \sin\alpha=\frac{5}{13}`

`\implies \cos\alpha=\frac{12}{13}`

The equation of line in the normal form is `x\cos \alpha+y\sin \alpha=p`

Here, `p=3`

Therefore, required equation of line is,
` frac{12}{13}x+frac{5}{13}y=3`

`\implies 12x+5y=39`