Question 24

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Jambmaths question: 

A straight line makes an angle of 30o with the positive x – axis and cut the y – axis at y =5. Find the equation of the straight line

Option A: 

$\tfrac{1}{10}x+5$

Option B: 

$y=x+5$

Option C: 

$\sqrt{3}y=-x+5\sqrt{3}$

Option D: 

$\sqrt{3}y=x+5\sqrt{3} $

Jamb Maths Solution: 

$\begin{align}  & slope=\tan \theta  \\ & m=\tan {{(180-30)}^{o}} \\ & m=\tan {{150}^{o}}=-\tan {{30}^{o}} \\ & m=-\tfrac{1}{\sqrt{3}} \\ & \text{when}\,x=0,\text{ }y=5 \\ & \text{Using one point slope form equation} \\ & (y-{{y}_{1}})=m(x-{{x}_{1}}) \\ & y-5=-\tfrac{1}{\sqrt{3}}(x-0) \\ & y-5=-\tfrac{x}{\sqrt{3}} \\ & \sqrt{3}y-5\sqrt{3}=-x \\ & \sqrt{3}y=-x+5\sqrt{3} \\\end{align}$

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