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

 Jun 17, 2021

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) 


⇒ \(x= {7\over 6}\)   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 laugh

 Jun 17, 2021
edited by amygdaleon305  Jun 17, 2021

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