a line drawn through `A(4,-1)` parallel to the line `3x-4y+1=0`. Find the coordinates of two points on this line which are at a distance of `5` units from `A`

Explanation

The equation of straight line passing through the point `A(4,-1)` parallel to the line `3x-4y+1=0` is,

`y+1=m(x-4)` where `m=` the slope of line `3x-4y+1=0` as both are parallel to each other.

`\implies y+1=3/4(x-4)`

`\implies 3x-4y-16=0`

Given `r=pm5`

Coordinates of required points are given by,

`frac{x−4}{4/5}=frac{y+1}{3/5}=r`
where r is the distance of any point on this line from the point A`(4,-1)`

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