Question on: JAMB Mathematics - 2001
A straight line makes an angle of 30° 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
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Correct Option: D
Cos 30 = 5/x
x cos 30 = 5, => x = 5√3
Coordinates of P = -5, 3, 0
Coordinates of Q = 0, 5
Gradient of PQ = (y2 - y1) (X2 - X1) = (5 - 0)/(0 -5√3)
= 5/5√3 = 1/√3
Equation of PQ = y - y1 = m (x -x1)
y - 0 = 1/√3 (x -(-5√3))
Thus: √3y = x + 5√3
x cos 30 = 5, => x = 5√3
Coordinates of P = -5, 3, 0
Coordinates of Q = 0, 5
Gradient of PQ = (y2 - y1) (X2 - X1) = (5 - 0)/(0 -5√3)
= 5/5√3 = 1/√3
Equation of PQ = y - y1 = m (x -x1)
y - 0 = 1/√3 (x -(-5√3))
Thus: √3y = x + 5√3
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