JEE MAIN - Mathematics (2002 - No. 11)
A triangle with vertices $$\left( {4,0} \right),\left( { - 1, - 1} \right),\left( {3,5} \right)$$ is :
isosceles and right angled
isosceles but not right angled
right angled but not isosceles
neither right angled nor isosceles
Explanation
$$AB = \sqrt {{{\left( {4 + 1} \right)}^2} + {{\left( {0 + 1} \right)}^2}} = \sqrt {26} ;$$
$$BC = \sqrt {{{\left( {3 + 1} \right)}^2} + {{\left( {5 + 1} \right)}^2}} = \sqrt {52} $$
$$CA = \sqrt {{{\left( {4 - 3} \right)}^2} + {{\left( {0 - 5} \right)}^2}} = \sqrt {26} ;$$
In isosceles triangle side $$AB=CA$$
For right angled triangle, $$B{C^2} = A{B^2} + A{C^2}$$
So, here $$BC = \sqrt {52} $$ or $$B{C^2} = 52$$
or $$\,\,\,\,\,\,\,\,\,{\left( {\sqrt {26} } \right)^2} + {\left( {\sqrt {26} } \right)^2} = 52$$
So, the given triangle is right angled and also isosceles.
$$BC = \sqrt {{{\left( {3 + 1} \right)}^2} + {{\left( {5 + 1} \right)}^2}} = \sqrt {52} $$
$$CA = \sqrt {{{\left( {4 - 3} \right)}^2} + {{\left( {0 - 5} \right)}^2}} = \sqrt {26} ;$$
In isosceles triangle side $$AB=CA$$
For right angled triangle, $$B{C^2} = A{B^2} + A{C^2}$$
So, here $$BC = \sqrt {52} $$ or $$B{C^2} = 52$$
or $$\,\,\,\,\,\,\,\,\,{\left( {\sqrt {26} } \right)^2} + {\left( {\sqrt {26} } \right)^2} = 52$$
So, the given triangle is right angled and also isosceles.
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