If on the coordinate plane (6, 2) and (0, 6) are the endpoints of the

Official Solution:

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?

A. \(\frac{1}{\sqrt{2}}\)
B. 1
C. \(\sqrt{2}\)
D. \(\sqrt{3}\)
E. \(2\sqrt{3}\)


Given endpoints of diagonal of a square: \(B(0,6)\) and \(D(6,2)\). Let other vertices be \(A\) (closest to the origin) and \(C\) (farthest to the origin):



Length of the diagonal would be: \(D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(6-0)^2+(2-6)^2}=\sqrt{52}\)

Coordinates of the midpoint \(M\) of the diagonal would be: \(M(x,y)=(\frac{6+0}{2},\frac{2+6}{2})=(3,4)\).

{The slope of line segment AM}*{The slope of line segment BD}=-1 (as they are perpendicular to each other), so \(\frac{y-4}{x-3}*\frac{6-2}{0-6}=-1\), which simplifies to \(y-4=\frac{3}{2}(x-3)\)

The distance between the unknown vertices to the midpoint is half the diagonal:

\((x-3)^2+(y-4)^2=(\frac{\sqrt{52}}{2})^2=13\);

\((x-3)^2+\frac{9}{4}(x-3)^2=13\);

\((x-3)^2=4\);

Which gives: \(x=1\) and \(y=1\) OR \(x=5\) and \(y=7\).

Hence point \(A\) is at \((1,1)\) and point \(C\) is at \((5,7)\). Closest to the origin is \(A\). Distance \(OA=\sqrt{2}\)


Answer: C

Similar question: https://gmatclub.com/forum/on-the-coord ... 27267.html

diagonal lenth ; √52
s√2=√52
s=√26
the nearest point on the coordinate (1,1) as we get s=√26
distance from origin ; 1
IMO B

Check these posts:

https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2013/04 ... ry-part-i/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2013/0 ... y-part-ii/

I was able to calculate the length of the side of the square = square root (26).

How do I get the coordinates of the vertex from this length? If I try to solve it using equation. I end up with a quadratic equation with x and y. Is there a technique I am missing?

The distance between the origin and the closest vertex constitutes the hypotenuse of a right triangle.
The answer choices represent the length of that hypotenuse.
Of the five answer choices, C seems the most likely option, since \(\sqrt{2}\) is the hypotenuse of a 45-45-90 triangle with sides 1, 1 and \(\sqrt{2}\).
Option C suggests the following figure:
In the figure above, the origin and point B form the hypotenuse of a 45-45-90 triangle with sides 1, 1 and \(\sqrt{2}\).
We can see that AB and BC form the vertex of a square.


Mathematically, we can prove that B is the desired vertex as follows:
AB and BC each form the hypotenuse a right triangle with legs 1 and 5, implying that AB=BC.
Slope of \(AB = \frac{6-1}{0-1} = -5\) and slope of \(BC = \frac{2-1}{6-1} = \frac{1}{5}\), implying that AB and BC are perpendicular, since their slopes are negative reciprocals.

We are given the 1st Diagonal in the Square as the Distance from Point (0,6) to Point (6,2)

Rule for Squares: the 2 Diagonals are Congruent, Perpendicular Bisectors of each other.

Rule in Coord. Geometry: the Perpendicular Bisector of a Line Segment will meet the Segment at the Midpoint and have a Slope that is the (-)Negative Reciprocal of the Slope of the Line Segment


Slope of the Diagonal Given = m1 = Rise/Run = (6 - 2) / (0 - 6) = 4 / -6 = (-)2/3

Mid-Point of the Diagonal = Point (3 , 4)

If from the Mid-Point (3 , 4) we move 2 Units DOWN on the Y-Axis and 3 Units RIGHT on X-Axis (following the Exact Units of the Slope) we end up at Point (6 , 2), which is the Given Vertex of the Square.

Similarly, if we Count 3 Units DOWN on the Y-Axis and 3 Units RIGHT on the X-Axis from the Other Given Vertex (0 , 6), we will arrive at the Mid-Point of the Diagonal (3 , 4)


Therefore, if we use the (-)Negative Reciprocal Slope and follow the Exact Units of the (-)Neg. Reciprocal Slope from the Mid-Point (3, 4) ----- we will be able to Find the 2 Other Vertices of the Square since the Diagonals are Congruent and All the Sides of the Square are Equal (basically, Symmetry)

The (-)Negative Reciprocal Slope of m1 = - (2/3) is:

m2 = + 3/2


Using the Exact Simplified Units of the m2 Slope:

Counting From the Mid-Point (3 , 4) UP 3 Units on the Y-Axis and RIGHT 2 Units on the X-Axis we find Vertex Point (5 , 7)

The 4th and Final Vertex Closest to the Origin can be found by Following the Slope in REVERSE. From the Mid-Point (3 , 4), counting DOWN 3 Units on the Y-Axis and to the LEFT 2 Units on the X-Axis, we find Vertex Point (1 , 1)


Lastly, the Distance from Origin (0 , 0) to Vertex (1 , 1) = sqrt { (1 -0)^2 + (1 - 0)^2 } = sqrt{2}

Answer is the Square Root of 2

-C-

Hello

My approach was a little different.
Since √2a= diagonal= √52 => side a= √26. Let the point we need be (x,y).
Now, as all sides of a square are equal, the distance from this point (x,y) to the vertices (0,6) and (6,2) must be equal.
√(x-0)^2 + (y-6)^2 = √(x-6)^2 + (y-2)^2.
Solving this gives us 3x-2y=1.
(1,1) is a solution for this equation. Distance between (0,0) and (1,1) is √2.
Of course, this single equation can have infinite solutions, but (1,1) keeps the distances on both sides equal. I cross-checked (1,1). Guess, I got lucky on my first try.
If any one could further elaborate on my idea and see if the hit and trial can be eliminated in order to arrive at (1,1), please do let me know.

Hope it helps!

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?

Let (x,y) be coordinates of endpoint of other diagonal.

\((x-6)^2 + (y-2)^2 = (x-0)^2 + (y-6)^2 = {(6-2)^2 + (0-6)^2}/2 = 26\)
\(x^2 - 12x + 36 + y^2 - 4y + 4 = x^2 + y^2 - 12y + 36\)
-12x + 4 = -8y ; 3x = 2y + 1
x = (2y+1)/3

\(x^2 + (y-6)^2 = 26\)
\((2y+1)^2/9 + y^2 - 12y + 36 = 26\)
\(4y^2 + 4y + 1 + 9y^2 - 108y + 90 = 0\)
\(13y^2 - 104y +91 = 0\)
\(13y^2 - 13y - 91y + 91 = 0\)
(13y-91)(y-1) = 0
y = 1 or y = 91/13 = 7

(x,y) = {(1,1),(5,7)}

Distance from point (0,0) to (1,1) = \(\sqrt{(1-0)^2 + (1-0)^2} = \sqrt{2}\)

IMO C