A) \[y=x+\sqrt{3}\]
B) \[\sqrt{3y}=x-\sqrt{3}\]
C) \[y=\sqrt{3}x-\sqrt{3}\]
D) \[\sqrt{3}y=x-1\]
Correct Answer: B
Solution :
Slope of \[x+\sqrt{3}y=\sqrt{3}\]is \[-\frac{1}{\sqrt{3}}={{m}_{1}}\](let) So, \[tan\text{ }\theta =-\frac{1}{\sqrt{3}}\] \[\theta =150{}^\circ \]You need to login to perform this action.
You will be redirected in
3 sec