Find the coordinates of the point on the line y = 7x + 3 that are closest to its origin.

Question

anonymous · Answer

work out x and y-intercepts

amistre64 · Answer

define "closest to its origin" please

amistre64 · Answer

i assume you want a perpendicular line that includes the point (0,0)

anonymous · Answer

oh yes - i assumed the points on the axes!!

amistre64 · Answer

then solve them in a system of equations that results in the common point

anonymous · Answer

perpendicular distance = 3/(50)^1/2

anonymous · Answer

i'm not sure better check it

amistre64 · Answer

we could do vector to maybe :)

anonymous · Answer

perpendicular satisfying 0,0

anonymous · Answer

closest to its origin as close as the "x values being a negative decimal not exceeding -3 and the y value can possible range until -17

amistre64 · Answer

so this slope is a "7" ; flip it spank it and throw it back down for the perp...

(-1/7)x and that should be it for that ...... but what does this new stuff mean?

anonymous · Answer

it was a hint at the answer. idk thats what i got

anonymous · Answer

I was given*

anonymous · Answer

the closest distance should be $$\frac{3}{\sqrt{50}} $$

amistre64 · Answer

im gonna assume its textbook noise then :)

y = 7x + 3
y = (-1/7)x ; multiply by 49

    y =   7x + 3
49y = -7x
---------------
50y = 3

y = 3/50  in this case ? maybe?

anonymous · Answer

-.42, .06

amistre64 · Answer

y = 7x + 3 ; multiply by -1
y = (-1/7)x

-y = - 7(7)/7 x   -3 
  y =   -1/7 x
--------------
0 =   -50/7  x  -3

x = -3(7)/50 = -21/50

amistre64 · Answer

:)  accursed fraction lol; but ill assume its right

anonymous · Answer

It is (: As I beat you to the second point :p

anonymous · Answer

http://www.intmath.com/plane-analytic-geometry/perpendicular-distance-point-line.php