Question

Geometry help needed please!

Answer 1

@SolomonZelman @zepdrix

Answer 2

You have the point \(\color{black}{\displaystyle (x,y)}\) given as, \(\color{black}{\displaystyle (x,y)=(-t+6,~3t-1)}\).
The distance from \(\color{black}{\displaystyle (x,y)}\) to \(\color{black}{\displaystyle (-2,0)}\) is: \(\color{black}{\displaystyle D(x,y)=\sqrt{(x+2)^2+(y-0)^2}}\).
Which is an equivalent to: \(\color{black}{\displaystyle D(t)=\sqrt{(-t+6+2)^2+(3t-1)^2}=(x,y)=(-t+6,~3t-1)}\)

Answer 3

oh excuse me, on the last line, the square root should be the last thing, ignore the rest of it.

Answer 4

\(\color{black}{\displaystyle D(t)=\sqrt{10t^2-22t+65}}\)

Answer 5

Then minimize this D(t), this will give you the value of t, for which (x,y) is closest to (-2,0). And then, all you have to do, is to plug this value of t that you found, into (x,y).