maxmin problem: point on -2x+2y-4=0 closest to (-1,2)

[kcbrat]

find the point on the line -2x+2y-4=0 which is closest to the point (-1,2).

help?

 

[stapel]

find the point on the line -2x+2y-4=0 which is closest to the point (-1,2).

Use the Distance Formula that you learned about back in algebra: plug the formula for (x, y) (by solving -2x + 2y - 4 = 0 for "y=") and the point into the Distance Formula, and find the minimizing value of x.

If you get stuck, please reply with a clear listing of your work and reasoning so far, starting with what you got when you solved the posted equation for "y=". Thank you!

Eliz.

 

[soroban]

Hello, kcbrat!

This can be solved without Calculus . . .

\(\displaystyle \text{Find the point on the line }L\!:\;-2x+2y-4\:=\:0\:\text{ which is closest to the point }P(-1,2).\)

\(\displaystyle \text{The line is }L\!:\;\;y \:=\:x + 2\text{, with slope 1.}\)

\(\displaystyle \text{The line }M\text{ through }P\text{ perpendicular to }L\text{ is: }\;y-2 \:=\:-1(x + 1) \quad\Rightarrow\quad y \:=\:-x + 1\)


\(\displaystyle \text{Now find the intersection of line }L\text{ and line }M.\)