dave13
generis
Bunuel
In the xy-plane, which of the following points is the greatest distance from the origin?
(A) (0,3)
(B) (1,3)
(C) (2,1)
(D) (2,3)
(E) (3,0)
Points on the axes lie on the same line as the origin, and have the same x- or y-coordinates. Subtract 0 from the non-zero coordinate.
For other points, you could construct a right triangle with horizontal and vertical legs that = length of coordinates.
(A) (0,3)
From y-coordinates of (0,3) and (0, 0)
Distance \(= (3 - 0) = 3\)
(B) (1,3)
Distance, d, from a right triangle with leg = 1 and leg = 3 (Pythagorean theorem):
\((1^2 + 3^2) = d^2\)
\((1 + 9) = 10 = d^2\)
Distance \(= \sqrt{10}\)
(C) (2,1)
Distance, d, from a right triangle:
\((2^2 + 1^2) = (4 + 1) = 5 = d^2\)
Distance \(= \sqrt{5}\)
(D) (2,3)
Distance, d, from a right triangle:
\((2^2 + 3^2) = (4 + 9) = 13 = d^2\)
Distance \(= \sqrt{13}\)
(E) (3,0)
From x-coordinates of (3,0) and (0,0)
Distance \(= (3 - 0) = 3\)
Answer D
Hi
generis 
its me again
i have a question
why didnt you put options A and E under radical sign square it
(E) (3,0)
From x-coordinates of (3,0) and (0,0)
Distance \(=\sqrt{(3 - 0)^2}= 3\)
the applies for option A no ?
merci beaucoup

dave13 , you are correct. You could use the distance formula for A and E.
I am not positive that is the question you ask.
Your "distance" for E seems to leave out part of the distance formula.
If it IS supposed to be the distance formula, I think it should say:
\(=\sqrt{(3 - 0)^2 +\\
(0 - 0)^2}= 3\)
Three comments.
1) I neither like nor use the distance formula, described as "the Pythagorean theorem in disguise."
I am happy with the unconcealed version.
So I draw right triangles and use the Pythagorean theorem to find or compare distances.
2) if you work well with the distance formula, use it!
3) regarding A and E, I will restate less cryptically than above:
"Points [that lie] on the [x- or y-] axes lie on the same line as [points on] the origin, and have [either] the same x- or [the same] y-coordinates. Subtract 0 from the non-zero coordinate."
Answer A's "two" points, (0,0) and (0,3), lie on the y-axis.
Their x-coordinates = 0. Graph it.
There is no need for the distance formula (or my right triangles).
The distance of point (0,3) to origin (0,0)?
\(y_2 - y_1 = 3\)Answer E's two points lie on the x-axis. Same analysis, except the y-coordinates = 0.
Subtract x-coordinates.
But if the distance formula is your go-to, again: use whatever works for you.
There is a saying in the U.S. (poor grammar, do not use in SC!):
If it ain't broke, don't fix it. 
De rien