Question

Consider the line y=1 x -1 and the point P=(2,0).

Answer 1

Answer:

Step-by-step explanation:

d(x)=√((x-2)2+(y-0)²)

=√((x-2)²+y²)

=√((x-2)²+(x-1)²)

=√(x²-4x+4+x²-2x+1)

=√(2x²-6x+5)

D=d²(x)=2x²-6x+5

$b.\\\frac{dD}{dx}=4x-6\\\frac{dD}{dx}=0,gives \\4x-6=0\\x=\frac{3}{2}\\\frac{d^2D}{dx^2}=4 >0 at x=\frac{3}{2}\\so~D~or~d^2~or~d~is~minimum~at~x=\frac{3}{2}\\so~y=1(\frac{3}{2} )-1=1/2=0.5\\so~nearest~point~is~(1.5,0.5)$