A) right angled triangle
B) isosceles triangle
C) equilateral triangle
D) scalene triangle
Correct Answer: B
Solution :
| Let \[A\left( -4,\,0 \right),\,B\left( 4,\,0 \right),\,C\left( 0,\,3 \right)\]are the given vertices. |
| Now, distance between A (-4, 0) and B (4,0), |
| \[AB=\sqrt{{{\left[ 4-\left( -4 \right) \right]}^{2}}+{{\left( 0-0 \right)}^{2}}}\] |
| \[\left[ \because \,dis\tan ce\,between\,two\,po\operatorname{int}s\,\left( {{x}_{1}},\,{{y}_{1}} \right) \right]\] |
| And \[\left( {{x}_{2}},\,{{y}_{2}} \right),\,d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\] |
| \[=\sqrt{{{\left( 4+4 \right)}^{2}}}=\sqrt{{{8}^{2}}}=8\]units |
| Distance between 5(4,0) and C(0, 3), |
| \[BC=\sqrt{{{\left( 0-4 \right)}^{2}}+{{\left( 3-0 \right)}^{2}}}\] |
| \[=\sqrt{16+9}=\sqrt{25}=5\] units |
| Distance between A(-4,0) and C(0, 3), |
| \[AC=\sqrt{{{\left[ 0-\left( -4 \right) \right]}^{2}}+{{\left( 3-0 \right)}^{2}}}\] |
| \[=\sqrt{16+9}=\sqrt{25}=5\]units |
| \[\because \,\,\,\,BC=AC\] |
| Hence, \[\Delta ABC\]is an isosceles triangle because an isosceles triangle has two sides equal. |
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