The houses of 4 friends are located at the points (6, 6), (0, 6), (3, 3) and (6, 0) as shown in the figure. A, C and B are the vertices of a right triangle. Find the distance between the houses at points C and D.

\sqrt{6} units

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\sqrt{3} units

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\sqrt{2} units

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\sqrt{5} units

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Solution

The correct option is C 3 $\sqrt{2} units$
$Distance between two points (x_{1}, y_{1}) and (x_{2}, y_{2}) is \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} ∴ Distance between the houses at points C and D = \sqrt{(6 - 3)^{2} + (6 - 3)^{2}} = \sqrt{9 + 9} = 3 \sqrt{2} units$

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The houses of 4 friends are located at the points (6, 6), (0, 6), (3, 3) and (6, 0) as shown in the figure. A, C and B are the vertices of a right triangle. Find the distance between the houses at points C and D.

The correct option is C 3√2unitsDistance between two points (x1,y1) and (x2,y2) is √(x2−x1)2+(y2−y1)2∴Distance between the houses at points C and D =√(6−3)2+(6−3)2 =√9+9 =3√2units

The correct option is C 3 $\sqrt{2} units$
$Distance between two points (x_{1}, y_{1}) and (x_{2}, y_{2}) is \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} ∴ Distance between the houses at points C and D = \sqrt{(6 - 3)^{2} + (6 - 3)^{2}} = \sqrt{9 + 9} = 3 \sqrt{2} units$