Find the distance of the point (2,5) from the line `3x+y+4=0` measured parallel to the line `3x-4y+8=0`

Explanation

Slope of `3x-4y+8=0` is `3/4` which imply `tan theta=3/4`

`implies cos theta=4/5` and `sin theta =3/5`

The equation of straight line passing through the point A`(2,5)` having slope `3/4`

`frac{x−2}{4/5}=frac{y−5}{3/5}=r`
where r is the distance of any point B on this line from the point A`(2,5)`

Any point B `=(2+frac{4r}{5},5+frac{3r}{5})` is on the line `3x+y+4=0`

`therefore3(2+frac{4r}{5})+(5+frac{3r}{5})+4=0`

`\implies6+frac{12r}{5}+5+frac{3r}{5}+4=0`

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