If on the coordinate plane (6,2) and (0,6) are the endpoints of the diagonal of a square, what is the distance between point (0,0) and the closest vertex of the square?
1/√2
1
√2
√3
2√3
Let us discuss the alternative methods for this question.
If on the coordinate plane (6,2) and (0,6) are the endpoints
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- richachampion
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OA: C
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The diagonals of a square bisect each other.richachampion wrote:If on the coordinate plane (6,2) and (0,6) are the endpoints of the diagonal of a square, what is the distance between point (0,0) and the closest vertex of the square?
1/√2
1
√2
√3
2√3
The point of intersection is equal to the MIDPOINT of each diagonal.
DRAW the given diagonal, along with its midpoint:
The diagonals of a square are perpendicular.
Draw a perpendicular line through the midpoint of the given diagonal:
In the figure above, the second diagonal seems to pass through (1,1).
Test whether (1,1) could be a vertex of the square, as follows:
If (1,1) is a vertex of the square, then the two sides of the square that intersect at (1,1) must be perpendicular.
The slopes of perpendicular lines are negative reciprocals.
Test whether (1,1) yields two sides whose slopes are negative reciprocals.
Slope of the side formed by (0,6) and (1,1) = (1-6)/(1-0) = -5.
Slope of the side formed by (6,2) and (1,1) = (1-2)/(1-6) = 1/5.
Success!
The resulting slopes are negative reciprocals.
Thus, the vertex closest to the origin is (1,1).
To determine the distance between (0,0) and (1,1), draw a right triangle:
The resulting triangle is a 45-45-90 triangle with a hypotenuse of √2.
Thus, the distance between (0,0) and the closest vertex = √2.
The correct answer is C.
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- richachampion
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Thank you so much for taking so much time in creating figures to make the solution succinct and lucid.GMATGuruNY wrote: The resulting triangle is a 45-45-90 triangle with a hypotenuse of √2.
Thus, the distance between (0,0) and the closest vertex = √2.
The correct answer is C.
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