Square

Not sure. What's the source of the question. This doesnt look like a gmat qn.

Anyways, Mid point of the diagonal is (3,2). Since, the diagonal/2 is 3 the other 2 points will be (3,-1) and (3,5). Now, the closest line will be the one between (0,2) and (3,-1). The equation for this line will be:

x + y + 2 = 0 (Solve for 2 points). Now, the closest distance from a point to a line will be the shortest distance, which is the perpendicular line.

If you have a point (x1,y1) and a line ax+by+c = 0, the perpendicular distance will be,

Modulus( (ax1+by1+c) / root(a^2 + b^2) ). Hence, this distance will be:
x1 = 0, y1 = 0. a = 1 b = 1 c = 2.

Ans: root(2).

enniguy wrote:Not sure. What's the source of the question. This doesnt look like a gmat qn.

Anyways, Mid point of the diagonal is (3,2). Since, the diagonal/2 is 3 the other 2 points will be (3,-1) and (3,5). Now, the closest line will be the one between (0,2) and (3,-1). The equation for this line will be:

x + y + 2 = 0 (Solve for 2 points). Now, the closest distance from a point to a line will be the shortest distance, which is the perpendicular line.

If you have a point (x1,y1) and a line ax+by+c = 0, the perpendicular distance will be,

Modulus( (ax1+by1+c) / root(a^2 + b^2) ). Hence, this distance will be:
x1 = 0, y1 = 0. a = 1 b = 1 c = 2.

Ans: root(2).

This is a from GMAT Clubs. Your answer is correct. Bit, since the end points of the diagonal are given, can we not draw a square directly? With the above solution, I tried, but unable to figure the square on coordinate plane. Also, I did not understand the below:

Since, the diagonal/2 is 3 the other 2 points will be (3,-1) and (3,5). Now, the closest line will be the one between (0,2) and (3,-1).

crackgmat007 wrote: This is a from GMAT Clubs. Your answer is correct. Bit, since the end points of the diagonal are given, can we not draw a square directly? With the above solution, I tried, but unable to figure the square on coordinate plane. Also, I did not understand the below:

Since, the diagonal/2 is 3 the other 2 points will be (3,-1) and (3,5). Now, the closest line will be the one between (0,2) and (3,-1).

We are figuring out points "as if" we have drawn the square without actually drawing it. You have been given two points of the diagonal (0,2) and (6,2). The distance is clearly 6. Midpoint of the sqaure will be diagonal/2. So, 3. On the co-ordinate system it will be (3,2). Now, the other diagonal will be vertical and of length 6. Hence, 3 on either side. Hence, (3,5) and (3,-1). Imagine a "Diamond" with center on (3,2)

Makes sense Thanks.

I did not understand the below, can you clarify pls:

x + y + 2 = 0 (Solve for 2 points). Now, the closest distance from a point to a line will be the shortest distance, which is the perpendicular line.

If you have a point (x1,y1) and a line ax+by+c = 0, the perpendicular distance will be,

Modulus( (ax1+by1+c) / root(a^2 + b^2) ). Hence, this distance will be:
x1 = 0, y1 = 0. a = 1 b = 1 c = 2.

crackgmat007 wrote:Makes sense Thanks.

I did not understand the below, can you clarify pls:

x + y + 2 = 0 (Solve for 2 points). Now, the closest distance from a point to a line will be the shortest distance, which is the perpendicular line.

If you have a point (x1,y1) and a line ax+by+c = 0, the perpendicular distance will be,

Modulus( (ax1+by1+c) / root(a^2 + b^2) ). Hence, this distance will be:
x1 = 0, y1 = 0. a = 1 b = 1 c = 2.

If you have 2 points, lets say (x1,y1) and (x2,y2), then the equation of the straight line passing through these two points will be:

(y2 - y1)/(x2 - x1) = (y-y1) / (x-x1).
Solving this equation for the two points (0,2) and (3,-1) wil give us the equation x+y+2 = 0.

Now we need to find the shortest distance between (0,0) and any vertex on the square. This line is the closest line of the square that passes near (0,0). So, if there is such a vertex (As per question), it should be on this line.

Now, the shortest distance between any point and a straight line is the perpendicular distance. How? Just google for it (https://www.mathopenref.com/coordpointdist.html ).

This distance can be calculated by substituting the co-ordinates of the point in the equation of the line correspondingly for x and y. (Look for the formula in my previous post).