A) equilateral
B) isosceles
C) right angle
D) None of these
Correct Answer: C
Solution :
| Let the points are \[A\left( 3,\,2 \right)\], \[B\left( -2,\,-3 \right)\]and C (2, 3). |
| Then, \[AB=\sqrt{{{\left( -2-3 \right)}^{2}}+{{\left( -3-2 \right)}^{2}}}\] |
| [\[\because\]distance\[=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}]\] |
| \[=\sqrt{{{\left( -5 \right)}^{2}}+{{\left( -5 \right)}^{2}}}=\sqrt{25+25}=\sqrt{50}\] |
| \[=7.07\] units (approx) |
| \[BC=\sqrt{{{\left( 2+2 \right)}^{2}}+{{\left( 3+3 \right)}^{2}}}=\sqrt{{{\left( 4 \right)}^{2}}+{{\left( 6 \right)}^{2}}}\] |
| \[=\sqrt{16+36}=\sqrt{52}=7.21\]units (approx) |
| and \[CA=\sqrt{{{\left( 3-2 \right)}^{2}}+{{\left( 2-3 \right)}^{2}}}\] |
| \[=\sqrt{{{\left( 1 \right)}^{2}}+{{\left( -1 \right)}^{2}}}=\sqrt{1+1}\] |
| \[=\sqrt{2}=1.41\] (approx) |
| Also, \[{{\left( \sqrt{52} \right)}^{2}}={{\left( \sqrt{50} \right)}^{2}}+{{\left( \sqrt{2} \right)}^{2}}\] |
| \[\Rightarrow \,\,B{{C}^{2}}=A{{B}^{2}}+C{{A}^{2}}\] |
| So, by converse of Pythagoras theorem, |
| \[\angle A=90{}^\circ\] |
| Hence, \[\Delta BAC\]is a right angled triangle. |
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