Find the point on the line -6x - 5y + 1 = 0 which is closest to the point (3,1).
+526 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 
