gmat1220 wrote:

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) 1/sqrt (2)

(B) 1

(C) sqrt (2)

(D) sqrt (3)

(E) 2*sqrt (3)

Sol:

Midpoint of two points \((x_1, y_1) \hspace{2} & \hspace{2} (x_2, y_2)\)

\(Midpoint(M_{x1},M_{y1})=(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})\)

Similarly, let's find the midpoint for \((0,6) \hspace{2} & \hspace{2} (6,2)\)

\(Midpoint(M_{x1},M_{y1})=(\frac{0+6}{2}, \frac{6+2}{2})= (3,4)\)

Distance between two points \((x_1, y_1) \hspace{2} & \hspace{2} (x_2, y_2)\)

\(Distance = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

Similarly, the length of the diagonal is the distance between two points \((0,6) \hspace{2} & \hspace{2} (6,2)\)

\(Diagonal = \sqrt{(6-0)^2+(2-6)^2} = \sqrt{36+16} = \sqrt{52}\)

Equation of a line passing through points \((x_1, y_1) \hspace{2} & \hspace{2} (x_2, y_2)\)

\(y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\)

Similarly, Equation of the line segment or diagonal BD passing through \((0,6) \hspace{2} & \hspace{2} (6,2)\)

\(y-6=\frac{2-6}{6-0}(x-0)\)

\(y-6=-\frac{2}{3}x\)

\(y=-\frac{2}{3}x+6\)

The slope of the diagonal \(BD=-\frac{2}{3}\)

Slope of the diagonal BD * Slope of the diagonal AC = -1 [Note: Both diagonals are perpendicular]

\(-\frac{2}{3} * Slope_{(AC)} = -1\)

\(Slope_{(AC)} = \frac{3}{2}\)

Equation of a line with slope \(\frac{3}{2}\) passing through point \((3,4)\)

\(y-y_1 = m(x-x_1)\)

\(y-4 = \frac{3}{2}(x-3)\)

\(y = \frac{3}{2}x-\frac{1}{2}\) -----------------1

Now, if we take \((3,4)\) as the midpoint of a circle with radius=half of diagonal of the square;

Then, \(radius = \frac{\sqrt{52}}{2}\)

Equation of the circle will be:

\((x-x_1)^2+(y-y_1)^2=(radius)^2\)

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

We can solve equation 1 and 2 to get the value for x and y because the circle will meet the diagonal AC at two points

\((x-3)^2+(\frac{3}{2}x-\frac{1}{2}-4)^2=(\frac{\sqrt{52}}{2})^2\)

\((x-3)^2(1+\frac{9}{4})=13\)

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

\(x=5 \hspace{2} & \hspace{2}x=1\)

When \(x=1\), \(y=\frac{3}{2}x-\frac{1}{2}=\frac{3}{2}-\frac{1}{2}=1\)

When \(x=5\), \(y=\frac{3}{2}*5-\frac{1}{2}=\frac{15}{2}-\frac{1}{2}=7\)

Thus, we found the other two vertices of the square viz. \((1,1) \hspace{2} & \hspace{2} (5,7)\)

Vertex \((1,1)\) is closest to \((0,0)\)

Distance between \((0,0) \hspace{2} & \hspace{2} (1,1)\)

\(Distance = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

\(Distance = \sqrt{(1-0)^2+(1-0)^2} = \sqrt{2}\)

Ans: "B"

Attachments

Nearest_Vertex_From_Origin.PNG [ 9.91 KiB | Viewed 3226 times ]

_________________

~fluke

GMAT Club Premium Membership - big benefits and savings