coolio
alright, so the disatnce formula is
[tex]D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
solve for y in the equation given on top
2x+y-10=0
y=10-2x
ok
so (x1,y1)=(0,0), the first point
(x2,y2)=(x,10-2x)
so subsitute to get
[tex]D=\sqrt{(x-0)^2+(10-2x-0)^2}[/tex]
[tex]D=\sqrt{(x)^2+(10-2x)^2}[/tex]
[tex]D=\sqrt{x^2+4x^2-40x+100}[/tex]
[tex]D=\sqrt{5x^2-40x+100}[/tex]
find the miniumum
we only need to minimize what is under the square root so
take the derivitive of 5x²-40x+100
that is 10x-40
it is 0 at x=4
when x<4, the derivitive is negative
when x>4, the derivitive is positve
goes from negative to positive
so it is a minimum
at x=4
find y
y=10-2x
y=10-2(4)
y=10-8
y=2
(4,2)
find min distance
[tex]D=\sqrt{(x-0)^2+(y-0)^2}[/tex]
[tex]D=\sqrt{(4-0)^2+(2-0)^2}[/tex]
[tex]D=\sqrt{(4)^2+(2)^2}[/tex]
[tex]D=\sqrt{16+4}[/tex]
[tex]D=\sqrt{20}[/tex]
[tex]D=2\sqrt{5}[/tex]
the point, P, is (4,2) and the minimum distance is 2√5
