A) Yes
B) No
C) Can't find
D) None of the above
Correct Answer: A
Solution :
| Let A = (1, -1), B = (5, 2) and C = (9, 5). |
| Then, \[AB=\sqrt{{{\left( 5-1 \right)}^{2}}+{{\left( 2+1 \right)}^{2}}}\] |
| [\[\because\] distance \[=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\]] |
| \[=\sqrt{{{\left( 4 \right)}^{2}}+{{\left( 3 \right)}^{2}}}=\sqrt{16+9}\] |
| \[=\sqrt{25}=5\] units |
| \[BC=\sqrt{{{\left( 9-5 \right)}^{2}}+{{\left( 5-2 \right)}^{2}}}\]] |
| \[=\sqrt{{{\left( 4 \right)}^{2}}+{{\left( 3 \right)}^{2}}}=\sqrt{25}=5\] units |
| And \[AC=\sqrt{{{\left( 9-1 \right)}^{2}}+{{\left( 5+1 \right)}^{2}}}\] |
| \[=\sqrt{{{\left( 8 \right)}^{2}}+{{\left( 6 \right)}^{2}}}=\sqrt{100}=10\] units |
| \[\because\] Here, \[AC=AB+BC\] |
| \[\therefore\] A, B and (7 are collinear points. |
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