passpadi

2001

Mathematics

JAMB

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

Correct Answer: √3y = x + 5√3

Explanation

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

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
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