A) Isosceles
B) Equilateral
C) Scalene
D) Right - angled
Correct Answer: B
Solution :
(b): \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}=xy+yz+zx,\] \[\Rightarrow \] \[2{{x}^{2}}+2{{y}^{2}}+2{{z}^{2}}=2xy+2yz+2zx\] \[\Rightarrow \]\[{{x}^{2}}+{{y}^{2}}-2xy+{{y}^{2}}+{{z}^{2}}-2yz+{{z}^{2}}+{{x}^{2}}-2zx=0\] \[\Rightarrow \]\[{{(x-y)}^{2}}+{{(y-z)}^{2}}+{{(z-x)}^{2}}=0\] \[\therefore \] \[x-y=0\Rightarrow x=y\] \[y-z=0\Rightarrow y=z\] \[z~-x=0\Rightarrow z=x\] \[\therefore \]\[x=y=z\] \[\therefore \] It is an equilateral triangle.
You need to login to perform this action.
You will be redirected in
3 sec
