A) Parallelogram
B) Triangle
C) Square
D) None of these
Correct Answer: A
Solution :
| Let the points are A (2, 3), B (3,4), C(5,6) and D (4, 5). |
| Then, by distance formula |
| \[AB=\sqrt{{{\left( 3-2 \right)}^{2}}+{{\left( 4-3 \right)}^{2}}}=\sqrt{{{\left( 1 \right)}^{2}}+{{\left( 1 \right)}^{2}}}\] |
| \[=\sqrt{2}\] units |
| \[BC=\sqrt{{{\left( 5-3 \right)}^{2}}+{{\left( 6-4 \right)}^{2}}}\] |
| \[=\sqrt{{{\left( 2 \right)}^{2}}+{{\left( 2 \right)}^{2}}}=\sqrt{4+4}\] |
| \[=\sqrt{8}=2\sqrt{2}\] units |
| \[CD=\sqrt{{{\left( 4-5 \right)}^{2}}+{{\left( 5-6 \right)}^{2}}}\] |
| \[=\sqrt{{{\left( -1 \right)}^{2}}+{{\left( -1 \right)}^{2}}}=\sqrt{2}\] units |
| and\[AD=\sqrt{{{\left( 4-2 \right)}^{2}}+{{\left( 5-3 \right)}^{2}}}=\sqrt{{{\left( 2 \right)}^{2}}+{{\left( 2 \right)}^{2}}}\] |
| \[=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}\]units |
| Here, \[AB=CD\]and \[AD=BC\] i.e. the opposite sides are equal. So, given points are vertices of a parallelogram. |
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