A) right angled
B) equilateral
C) isosceles
D) none of these
Correct Answer: C
Solution :
Given three straight lines \[x+y=0,\text{ }3x+y-4=0\] and \[x+3y-4=0\] to be sides of triangles. On solving, we get vertices of\[\Delta ABC\]are \[A(2,-2),B(1,1),C(-2,2)\]. \[\therefore \] \[AB=\sqrt{{{(1-2)}^{2}}+{{(1+2)}^{2}}}\] \[=\sqrt{1+9}=\sqrt{10}\] \[BC=\sqrt{{{(-2-1)}^{2}}+{{(2-1)}^{2}}}\] \[=\sqrt{9+1}=\sqrt{10}\] \[CA=\sqrt{{{(2+2)}^{2}}+{{(-2-2)}^{2}}}\] \[=\sqrt{16+16}=\sqrt{32}\] \[\therefore \] \[AB=BC\] \[\Rightarrow \] triangle is isosceles.
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