Find the equation of a line which is at a distance of `5` units from origin and the perpendicular from origin to this line makes an angle `alpha` from the positive direction of `X`-axis where `tan alpha=(4)/(3)`.

Options

A. ` 3x+4y=25` or `3x+4y+25=0`
B. `4x+3y=25` or `4x+3y+25=0`
C. `3x-4y=25` or `3x-4y+25=0`
D. `4x-3y=25` or `4x-3y+25=0`

Answer

`A`

Explanation

Here `p=5` units and `tan alpha=(4)/(3)`
`:. Sec^(2)alpha=1+tan^(2)alpha`
`=1+(16)/(9)=(25)/(9)`
`implies sec alpha=+-(5)/(3)`
`implies cos alpha=+-(3)/(5)`
and `sin alpha=-(sin alpha)/(cos alpha)*-cos alpha`
`=tan alpha*cosalpha`
`=(4)/(3)* (+-(3)/(5))=+-(4)/(5)`
`:. cos alpha=(3)/(5)` and `sin alpha=(4)/(5)`
or `cos alpha=-(3)/(5)` and `sin alpha=-(4)/(5)`
Therefore, equation of a line is
`x cos alpha+y sin alpha =p`
`implies (3x)/(5)+(4y)/(5)=5` or `(-3x)/(5)-(4y)/(5)=5`
`implies 3x+4y=25` or `3x+4y+25=0`