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m15q19: (6, 2) and (0, 6) are the endpoints of the diagonal [#permalink]

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21 Mar 2007, 15:05

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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?

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

i know the answer and have the explanation. Please explain your approach or logic, thanks.

The mid point between (6,2) and (0,6) is (3,4)
You know that the vertexes of the diagonal mentioned has distance of 3 in the x coordinate and distance of 2 in the y coordinate.
Since both diagonals must be perpendicular, using the perpendicular property (inverse of slope), it must be that those distances are reversed.

This means that from point (3,4), the new vertexes will have distance of 3 in the y coordinate and distance of 2 in the x coordinate.

So two other vertexes will be at:
(3-2, 4-3) = (1,1)
and
(3+2, 4+3) = (5,7)

The closer to the (0,0) is (1,1)
Distance between (0,0) and (1,1) = sqrt(2)

now we know that diagonal as calculated is \(\sqrt{52}\) we also know that diagonal of a square is a\(\sqrt{2}\) , where a is the side Therefore one side is \(\sqrt{26}\). Let C (x, y) be the closest vertice now all sides are equal in a square Therefore distance between C and (0,6) and C and (6,2) will be equal to\(\sqrt{26}\)

solve for x and y the following equations \((x-0)^2 + (y-6)^2 =26\)and \((x-6)^2 + (y-2)^2 = 26\) we get C= (1,1) hence the distance is \(\sqrt{2}\) ANS

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?

Given endpoints of diagonal of a square: B(0,6) and D(6,2). Let other vertices be A (closest to the origin, left bottom vertex) 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: \(M(x,y)=(\frac{6+0}{2},\frac{2+6}{2})=(3,4)\).

Slope of the line segment AM*Slope of the line segment BD=-1 (as they are perpendicular to each other) --> \(\frac{y-4}{x-3}*\frac{6-2}{0-6}=-1\) --> \(\frac{y-4}{x-3}=\frac{3}{2}\) --> \(y-4=\frac{3}{2}(x-3)\)

Distance between the unknown vertices to the midpoint is half the diagonal: \((x-3)^2+(y-4)^2=(\frac{\sqrt{52}}{2})^2=13\) --> \((x-3)^2+\frac{9}{4}(x-3)^2=13\) --> \((x-3)^2=4\) --> \(x=1\) or \(x=5\) --> \(y=1\) or \(y=7\)

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

Nice. That's probably the most easiest and efficient way to solve this problem. Kudos from me!

You can actually answer the question without calculating any distances at all (except at the very end), provided you have a good understanding of the meaning of perpendicular slopes. The background: perpendicular slopes are negative reciprocals. If one line has a slope of a/b, a perpendicular line has a slope of -b/a. Now, if a line has a slope of a/b, that is just the ratio of the 'vertical change' to the 'horizontal change' of the line as you move to the right; if a line has a slope of a/b, that means if you move b units to the right, the line will rise by a units. On a perpendicular line, to travel the same distance, you'd *reverse* the horizontal and vertical changes (because the slopes are reciprocals) and also reverse the direction of one of the two changes (because of the negative sign). That is, on a perpendicular line, we'd travel the same distance if we moved right by a units and moved down by b units, or if we moved left by a units and up by b units.

So, in this question, we have a square. The diagonals of a square are perpendicular, and meet at their midpoints. The midpoint of the diagonals in this question is (3,4). Moving from (0,6) to (3,4), we go right by 3 units and down by 2 units. On the perpendicular diagonal, starting from the midpoint, to travel the same distance we need to either go right by 2 units and up by 3 units (to get to (5,7)), or left by 2 units and down by 3 units (to get to (1,1)). So the closest vertex to the origin is (1,1) and the distance is sqrt(2). _________________

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You don't need to use the distance formula here; instead we can use slopes. I posted this solution a while back to another forum:

We have two endpoints of a diagonal of a square. We can use the following:

-the midpoint of one diagonal is the midpoint of the other diagonal; -the diagonals are perpendicular.

If (0,6) and (6,2) are endpoints, (3,4) is the midpoint.

From (0,6) to (3,4), we go right 3 and down 2; that is, we increase x by 3 and decrease y by 2: the slope is -2/3. Consider the perpendicular diagonal- its slope is the negative reciprocal, i.e. 3/2. From (3,4), on a perpendicular line, to find a point the same distance from (3,4) as (0,6) is, we can decrease x by 2 and decrease y by 3, or we can increase x by 2 and increase y by 3. The endpoints of the other diagonal are (1,1) and (5,7).

The distance from the origin to (1,1) is sqrt(2). _________________

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it is clear that the closest vertex is the bottom left one (the bottom right is (6,2) and the upper left is (0,6))
so let's find its coordinates:

first, let's find the middle point (which is half way the diagonal)
since the whole diagonal spans between 0 and 6 on the x-axis the middle point is on 3. and on the y-axis it spans from 2 to 6 so the middle-point is at 4. so the middle point is at (3,4)

the bottom left is 90 degress down and left from the middle point, in the same distance of the end points of the diagonal....
now - here is a nice method to go 90 degrees from a given 2 points:
take the y difference of the given points, and make it the x difference, and vice versa, then go to the right direction:

so the x difference between (6,2) and (3,4) is 3 and y difference is 2
we our target point is with x-difference 2 (from (3,4)) and since it must be on the left - the x axis is 1.
the y difference is 3, and since the point should be on the bottom we substract it and get 1.

so the bottom left vertex is (1,1), and the distance from (0,0) is sqrt(2)

Here you go - The property of square is - diagonals intersect at midpoint and are equal in length. [(6, 2) and (0, 6)] is one diagonal. The mid point of this diagonal is M (6/2,(6+2)/2) i.e M (3,4)

The second diagonal passes through the nearest vertex (lets say A) and point M. Extend the second diagonal AM to meet the origin O(0,0). We have now OM.

OM = OA + AM. We need OA - distance of nearest vertex from the origin AM = Half of diagonal = sqrt(6^2 + 4^2)/2 = sqrt(52)/2 = sqrt(13) OM = sqrt(4^2 + 3^2) = 5 OA = OM - AM = 5 - sqrt(13) = 5 - 3.6 = 1.4 (approx). Hence C . sqrt(2) = 1.4

Slope of given diagonal =(6-2)/(0-6)=-2/3
Hence Slope of other diagonal =3/2
Midpoint of 'other dia' = (3,4)

Hence Slope=3/2 = (y-4) /(x-3)--------(1)--> Mavery see below corrected

distance of vertex (x,y) from (3,4) = sqrt 52/2
i.e (x-3)^2+ (y-4)^2= 52/4=13-----(2)

Let (x-3) =a,(y-4) =b
so eqn (2) becomes a^2+ b^2= 13
Dividing by b^2 throughout
(a/b)^2+ 1 =13/(a^2)
From eqn 1 a/b =3/2
Gives b=+/- 3
i.e y-4 = 3 =====> y= 7
y-4=-3======> y= 1
Similarly a = +/-2
====> x is 5,1
Vertices are (5,7) and (1,1)
Now we are having the two vertices ...clearly the closest one to origin is (1,1) ...
So distance is sqrt(2)...

What is the source of this Q ...?

Last edited by suithink on 22 Mar 2007, 10:43, edited 3 times in total.

Considering the 2 given endpoints of the diagonal of a square, find the other 2 vertices which doesn't spoil the geometry of figure which can only be done if we subtract 5 units on both x and y axis from the 2 given endpoints (6,2) and (0,6). And hence the vertex (6-5,6-5) = (1,1).

The other 2 vertices will be (1,1) and (5,7). The four vertices of square will be (1,1), (6,2), (5,7) and (0,6). The length of each side and digonals is same considering these vertices.

(1,1) is the vertex of the square closest to the origin and distance between (1,1) and origin is goona be Square Root of 2. _________________

Say (a,b) is a vertex of that square that does not makes the end points of the diagonal of that square.

Then, distances to the end points must be exactly 1/srqt2 of the lenght of the diagonal ( diagonal of a square and two sides of the square makes a right triangle)

lenght of diagonal is sqrt52 ( sqrt ((6-0)^2+(6-2)^2) ) so length of the edges of square is sqrt26.

Thus, lets calculate the distance of (a,b) to end points.

(a-0)^2 + (b-6)^2 = 26 (a-6)^2 + (b-2)^2 = 26

SOlving these,

(a,b) can be (1,1) or (5,7)

(1,1) has the shortest distance to (0,0) that is sqrt2.

Beauty of Graphs - Example 2 + Stunning Symmetry of Squares

If AC is the given diagonal with co-ordinates (0, 6) and (6, 2), we see it is a sloping line. The center point of the line is (3, 4) obtained by averaging x and y co-ordinates as given: (0 + 6)/2 = 3 and (6 + 2)/2 = 4. Think of a horizontal line PQ whose y co-ordinates were shifted by 2 units up and 2 units down to obtain AC

Attachment:

Graph1.jpg [ 9.53 KiB | Viewed 3975 times ]

Diagonal BD will be perpendicular to AC and will pass through (3, 4). It will have a corresponding line RS whose x co-ordinates will be shifted by 2 units right and 2 units left to obtain the co-ordinates of BD. (This is so because y co-ordinates of PQ were shifted by 2 units to obtain AC and PQ is perpendicular to RS)

Attachment:

Graph2.jpg [ 12.98 KiB | Viewed 3974 times ]

The closest co-ordinate to (0, 0) is (1, 1) and its distance from center is \(\sqrt{2}\) _________________

I use a method that is out of scope of the GMAT... I prefer to use the properties of perpendicular vectors...

First of all, I search the mid point I that is the center of gravity of the square.

I((6+0)/2 , (2+6)/2) <=> I(3,4)

Lets call:
o A : the vertice at (6,2)
o C : the vertice at (0,6)
o B : the vertice "above" the line AC
o D : the vertice "below" the line AC

Then, I calculate the vectors that we can have:
o Vector(IA) = (6-3,2-4) = (3,-2)
o Vector(IC) = (0-3,6-4) = (-3,2)

Then, as Vector(IA) is perpendicular to Vector(IB) and Vector(ID) is perpendicular to Vector(IC), we have the rules of perpendicular vectors:
o Vector(IB) = (-y(IA) , x(IA)) = (2,3)
o Vector(ID) = (-y(IC) , x(IC)) = (-2,-3)

Then, we rebuild the coordonate of B and D:
o B(3+2 , 4+3) <=> B(5 , 7)
o D(3-2 , 4-3) <=> D(1 , 1)

So, we have
o A(6,2)
o B(5,7)
o C(0,6)
o D(1,1)

Thus, OD is the shortest distance and OD = sqrt(2)

Let the nearest vertex is (x,y)
the length of the diagonal is = distance between (0,6) and (6,2) = sqrt(52). Hence the length of each side of the square = sqrt(26)
Now the distance between the each of the known vertices and the nearest vertex should be the length of a side. Equalise them will end up in the equation: 3x-2y =1
distance between (0,6) and (x,y) is sqrt(26), solving this y = 1 or 7. Putting the values of y in 3x-2y =1 , possible values of x are 1, 5.
Hence the nearest vertex is (1,1).
distance between (0,0) and (1,1) is sqrt(2).

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?

mid point of (0,6) and (6,2) is .. (3,4) and slope is -2/3 slope of perpendicular line passing through (3,4 ) 3/2 .=y2-y1/x2-x1 we can find two vertices by decreasing (or increasing) x by 2 and y by 3 other two points are (1,1) and (5,7)

(1,1) is nearest point to (0,0) and distince is sqrt(2).

I took 90 secs . _________________

Your attitude determines your altitude Smiling wins more friends than frowning

I spent an exorbitant amount of time on this problem.

Tried to tackled it via algebra, and got stuck.

This is what I did: . midpoint= (3,4) . slope= -2/3 . then perpendicular slope= 2/3 . plugging in (3,4) into y=3/2x+k got k=-1/2 -> y=3/2x-1/2 .....(eq1) . distance between points (diagonal)= sqrt(6^2+4^2)= sqrt(52) . half the distance= sqrt(13) . then distance between midpoint and vertex is given by: sqrt[(x-3)^2+(y-4)^2]=sqrt(13)->(x-3)^2+(y-4)^2=13 ....(eq2) . plugging equation1 into equation2 I tried to solve to get vertex (1,1) as said by other posters . but did not succeed. Reached x^2-3x+5=0, which has no real roots.

Good news is that googling around I found a fantastic shortcut:

From (3,4), on a perpendicular line (with slope 3/2), to find a point the same distance from (3,4) as (0,6) is, we decrease x by 2 and decrease y by 3, thus getting vertex (1,1), or we increase x by 2 and increase y by 3, thus getting vertex (5,7).

Some questions can be solved over 2 minutes. There are 75 minutes and 37 questions. Some can be done under 1 minute even under 30 seconds. Today I scored 51q and although 2 problems got my 6-7 minutes I finished exam in 64 minutes That will give help you I think. So the easier under 1 that will give you a chance to solve hards in 5-6 minutes.