Which of the following equations represents the graph in the figure?

A. $x-y=0$
B. $x+\sqrt{3}y=0$
C. $x-\sqrt{3}y=0$
D. $\sqrt{3}x+y=0$

Question

Which of the following equations represents the graph in the figure?

A. $x-y=0$
B. $x+\sqrt{3}y=0$
C. $x-\sqrt{3}y=0$ 
D. $\sqrt{3}x+y=0$

anonymous · Answer

|dw:1416724116234:dw|

ganeshie8 · Answer

it has to be of form `y = mx` because it is passing thru origin (0, 0)

anonymous · Answer

i forgot the equation of tangent....

ganeshie8 · Answer

slope = tan(30)

anonymous · Answer

oh 
$$y=\frac{ 1 }{ \sqrt{3}} x$$

anonymous · Answer

if yu look at your special angles in trigonometry tan(30) will be 1/sqrt(3)

anonymous · Answer

Therefore. the answer is C, right?

anonymous · Answer

Take a close look at given graph and choices you have.

Because slope is positive, we know that it have to be in form ` y=mx `, where m is positive, so it should be in form ` y-mx=0 ` or ` y-mx=0 `

We know it's not A because otherwise angle is supposed to be $45^o$

And we know it's not B nor D, so we are left with C.

anonymous · Answer

or ` mx-y=0 `***