Find the equation of a line whose perpendicular disatnce from the origin is `sqrt8` units and the angle between the positive direction of the x-axis and the perpendicular is `135^(@)` .

Explanation

Here `p=sqrt8` units are `alpha=135^(@)`
So, the equation of the given line in normal form is
`x cos alpha+y sin alpha=p`
`"where "alpha=135^(@) and p=sqrt8"units"`
`Leftrightarrow x cos135^(@)+y sin135^(@)=sqrt8`
`Leftrightarrow x cos135^(@)+y sin135^(@)=sqrt8`
`Leftrightarrow x((-1)/(sqrt2))+y((1)/(sqrt2))=sqrt18`
`[therefore cos135^(@)=cos(180^(@)-45^(@))=-cos45^(@) sin135^(@)=sin(180^(@)-45^(@))=sin45^(@)]`
`-x+y=4 Leftrightarrow x-y+4=0,` which is the required equation.