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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. \(2\sqrt{3}\)

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 \(OA=\sqrt{2}\)

Is there any shorter way to solve this problem. I have understood the solution but this seems to be a very tough and complex problem. I am wondering if this can be solved in some other way.
_________________

Hi akhil911, maybe this will help. I calculated the midpoint in the same way as described by Bunuel to find (3,4). Then I determined the delta from point B to the midpoint M as (0,6) - (3,4) = (3,2). In order to find point A simply swap the x and y and subtract from the midpoint: (3,4) - (2,3) = (1,1). Which gives √2 as distance.

Or for points (6,2) and (0,6) in this case: Absolute difference for x is 6-0=6 Absolute difference for y is 6-2=4 Using logic, since this is a square, sides need to be equal and average of 6+4 is 5 Hence, absolute difference of 5 units away is for x is x=1 and for y is y=1, with intersection at point (1, 1) being equidistant to B and D Then work out the distance as before

Shorter solution if one can draw a correct sketch of the figure:

The shortest distance from (0,0) to one of the vertexes of the square is perpendicural distance to A. A and (0,0) forms a square with s=1. The shortest distance will be the diagonal of that square which is V2s or 1 x V2 = V2 the answer.

Even if the graph is drawn somewhat right, we are looking for the answer that has nV2 in it.
_________________

Please kindly +Kudos if my posts or questions help you!

rachitshah It seems like the steps from the given explanation accomplish the following: 1) Find the distance of the original diagonal: (6-0)^2 + (2-6)^2 = X^2; X = root 52 2) Find the midpoint of the original diagonal: (6+0)/2, (6+2)/2 = (3,4) 3) Find the equation for the line for the other diagonal, since this will be the line that has the coordinates closest to (0,0) Key to this step: This line is the perpendicular bisector of the diagonal (the diagonals of squares are perpendicular bisectors = fact in geometry) You can find the equation for the line the other diagonal by doing the following: a) y = mx + b (generic equation for line) b) find the slope of the original diagonal: (6-2)/(0-6) = -2/3 and take the negative reciprocal of this = 3/2 since the other diagonal is perpendicular c) now you have y = 3x/2 + b; find the value of b by plugging in (3,4) and you get y = 3x/2 - 1/2, which is the same thing as y-4 = 3/2(x-3) 4) Distance from farthest coordinates to midpoint of diagonal = Distance from shortest coordinates to midpoint of diagonal = 1/2 the length of the diagonal (in this case, (root52/2)^2). 5) The distance from the farthest or shortest coordinates of the diagonal to the midpoint can be solved by using pythagorean theorem: (x-3)^2 +(y-4)^2 = (root52/2)^2. (root52/2)^2 = 13 6) From step 3 you know that y-4 = 3/2(x-3). Plug this into y for step 5 above ==> (x-3)^2 + (y-4)^2 = 13 --> (x-3)^2 + [3/2(x-3)]^2 = 13 --> (x-3)^2 +9/4(x-3)^2 = 13; solving for x gives you (x-3)^2 = 4, so x can be 1 or 5.

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

(x−3)2+(y−4)2=(52√2)2=13(x−3)2+(y−4)2=(522)2=13;

I don't understand what this step is doing?

This step is finding the distance from vertex A(x,y) to mid point M(3,4) or the length of AM you can say. since AC is also a diagonal, AM will be half its length (which is\sqrt{52} and diagonals of a square are equal in length.)

Hi akhil911, maybe this will help. I calculated the midpoint in the same way as described by Bunuel to find (3,4). Then I determined the delta from point B to the midpoint M as (0,6) - (3,4) = (3,2). In order to find point A simply swap the x and y and subtract from the midpoint: (3,4) - (2,3) = (1,1). Which gives √2 as distance.

Can you pls explain the logic behind this approach. What do you mean by delta here.

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

Hi Bunuel,

Is there any shorter way to solve this problem. Also, can you pls explain the logic behind the Skytail's approach in one of the above posts.

The midpoint of the diagonal is (3,4). The vector from the right bottom corner to the midpoint is V1 = <-3, -2>. You need to find a vector perpendicular to this vector and of the same magnitude and add it to the midpoint position vector to find the corners.

Vectors perpendicular meet:

-3x - 2y = 0 or y = (3/2) x So the form is <x, 3/2x>

The two perpendicular vectors that have the same magnitude as V1 will meet:

x^2 + [(3/2)x]^2 = 13 x = +/- 2

So the two vectors are <2,3> or <-2,-3>

To find the corner closest to the origin add <-2,-3> to the position vector of the midpoint <3,4> to get <1,1>.

The magnitude of this vector sqrt(2) is the answer.