On the coordinate plane (6, 2) and (0, 6) are the endpoints of the dia

can you explain in detail please!!

Question asks about the corners of the square. A line joining the corner of the diagonal to an adjacent corner must be the side of the square.

So the options given must first satisfy the criteria of being a corner of the given square. Only option C satisfies the criteria.

I used slightly different approach, but it seems like the one Vyshak meant to describe.

1. Draw the diagonal.
2. You see that point (0;2) cant be the corner, otherwise it will be not a square. So the corner cant have coordinates like (0;y) or (x:0). So only answers C and E is out.

If you can not see it right away then after drawing the diaganal, go to point 3 right away:
3. tTry answers and see whether the point can give you a squire
For example answer A:
draw point and connect corners of our squire. you see that one side from the left is 5, but the other side from the bottom is not 5 clearly.
The same case with answers B D E

If you need detailed explanation about answers pls let me know. I would be glad to help

This question is possibly designed to mislead test takers into thinking that sides of this square must be perfectly straight. Actually, we can also have a tilted square. The most natural solution would be to think that (0,2) must be the closet point to the origin however that is actually incorrect- it's nearly optical illusion- mathematically it could not be (0,2) because then one of the side lengths would be 4 (6-2 =4) and the other side length would be 6. This square is actually tilted- but more to the point, this question seems to loosely imply that these two endpoints that form a diagonal are also vertices. If that is true- then the point closest to the origin must be equidistant (equally distant) from both of those coordinates. Apply the distance formula



Thus
"C"

Bunuel it seems somewhat reasonable that this question is suggesting the two endpoints of the diagonal of the square must also be vertices- in which case the point that is closest to the origin must be equally distant. However, can we not actually have a diagonal that cuts through a square but does not necessarily connect two vertices? Kudos for explanation.

Not sure I can follow you...

The endpoints of the diagonal of a square and vertices of a square are the same thing - 2 diagonals, each has two endpoints = 4 vertices.

Clearly C

Just put the values to get slopes m1 and m2, andwe know m1 x m2 =-1
lets us put values for option C for m1 and m2
m1=(6-1)/(0-1) = -5
m2= (1-2)/(1-6) = 5
clearly m1 x m2 = -1
since we are multiplying slopes of side of square... these values are satisfying here so it is the shortest way.

Distance between co-ordinates (0,6) and coordinate we are going to find out = Distance between co-ordinate (6,2) and co-ordinate we are going to find out.

Only C satisfies this: \sqrt{\((0-1)^2 + (6-1)^2\)} = \sqrt{\((1-6)^2 + (1-2)^2\)}

My take at this

Distance will be root 52, now if i rotate the figure and ensure that i get a point which can give me a sum of diagonals = root 52, i should be good(since diagonals of a square are concurrent )

How can we get a 52
52 = 36 +16
36 = 7-1 8-2, we can eliminate A.
16 = 5-1 6-2, we can eliminate B, D

Out of C and E, which is the nearest point and maintain the figure’s consistency??

C

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Similar question: https://gmatclub.com/forum/if-on-the-co ... 07098.html

From the given points, we can get the equation of the diagonal as
2x+3y=6 and mid point as (3,4)

From this we can derive the equation of the other diagonal

This should be 3x-2y=1

Substitute the answer choices to see which one satisfies 3x-2y=1

Answer is C

Posted from my mobile device

Rules:

1- Diagonals of a Square are Perpendicular Bisectors - they will Intersect at the MID-POINT of Each Diagonal
2- The slopes of 2 Diagonal Lines that are Perpendicular --- m1 * m2 = (-)1 because the Slopes of Perpendicular Lines are Negative Reciprocals

If Point (0 , 6) and (6 , 2) make up 2 of the Vertices of the Square (they must because the Diagonal Connects the 2 Vertices) then we can first find the Slope of this Diagonal 1 = d1

Slope of d1 = (6 - 2) / (0 - 6) = - 2/3

The Mid-Point of the Diagonal 1 (d1) is at Point (3,4)

Diagonal 2 (d2) will Intersect d1 at this MID-POINT of Point (3,4). Further, d2's slope will be + 3/2 (the Negative Reciprocal of d1's slope - 2/3)

Following Diagonal 1's Slope, from the Vertex (0,6) to the Mid-Point Intersection at (3 , 4), you count DOWN 2 Units and to the RIGHT 3 Units

The Upper Right Corner Vertex of the Square will be Symmetrical from the Mid-Point. Using d2's Slope = 3/2 ----- we can count UP 3 and to the RIGHT 2 Units to get to the Vertex (5 , 7)

Similarly, following d2's Slope of 3/2, you can get to the Lower Left Corner Vertex. From the Mid-Point (3, 4) we count DOWN 3 Units and to the LEFT 2 Units to find Vertex (1 , 1)

This Vertex (1 , 1) will be closest to the Origin and is the Answer. C

IMO C

To find: Closest point, say B(x,y)
Given: It's a square, A(6,2) , C (0,6)

Distance A(6,2) & B(x,y) = Distance B(x,y) & C (0,6)

Test choices

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