A straight line makes an angle of 30^\circ with the positive x-axis and cuts the y-axis at y = 5. Find the equation of the straight line.
A
y = (x/10) + 5
B
y = x + 5
C
√3y = - x + 5√3
D
√3y = x + 5√3
Verified Correct Answer: D.
Step-by-Step Mathematical Solution
Cos 30 = \(\frac{5}{\text{x}}\)
x cos 30 = 5, => x = 5\(\sqrt{3}\)
Coordinates of P = - 5\(\sqrt{3}\), 0
Coordinates of Q = 0, 5
Gradient of PQ = (y\(_2\) - y\(_1\)) (X\(_2\) - X\(_1\)) = \(\frac{(5 - 0)}{(0 - 5\sqrt{3}}\))
= \(\frac{5}{5\sqrt{3}}\) = \(\frac{1}{\sqrt{3}}\)
Equation of PQ = y - y\(_1\) = m (x -x\(_1\))
y - 0 = \(\frac{1}{\sqrt{3}}\) (x -( -5\(\sqrt{3}\)))
Thus: \(\sqrt{3}\)y = x + 5\(\sqrt{3}\)
Source: Myschool Past Questions Archive (Record #1515)