Find the equation of a line passing through the origin so that the tangent of the angle between the line in quadrantI and the positive x-axis is sqrt3/3.
A. y=sqrt3/3x
B. y=sqrt3x
C.y=x
D.y=sqrt2/2x
thnks alot:)
Hint: The slope is (sqrt3)/3 and the y-intercept is 0.
Do the rest
so is it A??
yes!
To find the equation of a line passing through the origin with a given tangent angle, we can apply the trigonometric definition of the tangent function.
The tangent of an angle in a right triangle is defined as the ratio of the length of the side opposite the angle to the length of the side adjacent to the angle. In this case, the tangent of the angle between the line in the first quadrant and the positive x-axis is given as √3/3.
Let's assume the slope of the line passing through the origin is m. Since the line passes through the origin, we can write the equation in slope-intercept form as y = mx.
Now, we need to determine the value of m that satisfies the given tangent angle. We know that the tangent of an angle theta can be expressed in terms of the slope m using the formula tangent(theta) = m.
Given that tangent(theta) = √3/3, we can equate this to the slope m:
m = √3/3.
Therefore, the equation of the line passing through the origin is y = (√3/3)x.
From the given options, the equation that matches this expression is option A:
A. y = (√3/3)x.
So, the correct answer is A. y = (√3/3)x.