Find the point on the line -6x - 5y + 1 = 0 which is closest to the point (3,1).

Guest Jun 17, 2021

#1**+2 **

I'd written quite a lengthy answer to this but then the page accidently reloaded and damnit there goes efforts :/ but thankfully I'm gifted with extraordinary patience so I decided to give it another go.

So the point on the line \(-6x-5y+1=0\) which is closest to the point (3,1) is the perpendicular/normal on that line from that particular point, as we know that the normal is the shortest length from any point right?

Equation of line is

\(-6x-5y+1=0\) **...(1)**

Slope \(={-5\over 6}\)

Slope of normal \(={-1\over {-5\over 6}} = {6\over 5}\)

Now we find the equation of that normal from the point (3,1)

\(y-1={6\over 5}(x-3)\)

\(5y-5=6x-18\) ⇒\(6x - 5y-13=0 \) **...(2) **

Now that point is where the two lines intersect so we equate eq(1) and (2)

\(-6x-5y+1=6x-5y-13\)

⇒ \(x= {7\over 6}\) and \(y={-6\over 5}\)

**∴The point is \(({7\over 6},{-6\over 5})\)****. **

To wrap things up, I've also enclosed a graph here

I hope its clear.

P.S. Turns out, I've explained this answer lot better than the previous one lol. If anything I've gained from writing this answer twice, its that I've gained a lot more patience that I already have

amygdaleon305 Jun 17, 2021