Find the distance of the point `(3, 5)` from the line `2x + 3y = 14` measured parallel to the line `x-2y = 1.`

Explanation

Slope of `x-2y=1` is `1/2` which imply `tan theta=1/2`

`implies cos theta=2/sqrt5` and `sin theta =1/sqrt5`

The equation of straight line passing through the point A`(3,5)` having slope `1/2`

`frac{x−3}{2/sqrt5}=frac{y−5}{1/sqrt5}=r`
where r is the distance of any point B on this line from the point A`(3,5)`

Any point B `=(3+frac{2r}{sqrt5},5+frac{r}{sqrt5})` is on the line `2x+3y=14`

`therefore2(3+frac{2r}{sqrt5})+3(5+frac{r}{sqrt5})=14`

`\implies7+frac{7r}{sqrt5}=0`

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