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"
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~fluke
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