A) equilateral
B) isosceles
C) right angled
D) None of these
Correct Answer: B
Solution :
Given, lines are \[x+y=0,3x+y=4,x+3y=4\] Solving these equations, we get the vertices of the\[\Delta \]ABC as\[A(2,-2),B(1,1)\]and\[C(-2,2)\] \[\therefore \]\[AB=\sqrt{{{(1-2)}^{2}}+{{(1+2)}^{2}}}=\sqrt{1+9}=\sqrt{10}\] \[BC=\sqrt{{{(-2-1)}^{2}}+{{(2-1)}^{2}}}=\sqrt{9+1}=\sqrt{10}\] and\[CA=\sqrt{{{(2+2)}^{2}}+{{(-2-2)}^{2}}}=\sqrt{16+16}=\sqrt{32}\] \[\because \] \[AB=BC\] Hence, triangle is isosceles.
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