+0

Coordinates

0
41
1

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

Jun 17, 2021

1+0 Answers

#1
+505
+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

Jun 17, 2021
edited by amygdaleon305  Jun 17, 2021