Question

Find the slope of the line
i) Which bisects the first quadrant
ii) Which makes an angle of ${{30}^{\circ }}$ with the positive direction of y-axis measured anti-clockwise

Answer 1

Hint: Use the fact that the slope of the line can also be represented as the tangent of the angle which the line makes with the positive x-axis when going anticlockwise from the x-axis.
The value of m gives the slope of the line and then equate it to the tangent of the angle which the line makes with the positive x-axis when going anticlockwise from the x-axis as follows
$m=\tan \theta$
(Where $\theta$ is the angle that the line makes with the positive x-axis when going anticlockwise from the x-axis and m is the slope of the line which is inclined to the x-axis with the mentioned angle)
Complete step-by-step answer:
Now, in this question, for the first part, we can see that one quadrant consists of ${{90}^{\circ }}$ , so, the line which will bisect the first quadrant will make an angle of ${{45}^{\circ }}$ with the x-axis when measured anticlockwise.
For the second part, we can use the fact that if a line makes some angle the y-axis measured anti-clockwise and if ${{90}^{\circ }}$ is added to that angle , then the angle that we get is the angle that the same line makes with the x-axis when measured anti-clockwise.
As mentioned in the question, we have to find the slope of the line which is according to the information that is provided in the question.
i) A line which bisects the first quadrant
Now, as mentioned in the hint, we can use the information as follows
As the first quadrant comprises of ${{90}^{\circ }}$ , hence, the angle of the line that bisects it should be ${{45}^{\circ }}$ with the x-axis when measured anticlockwise and we can write it as follows

& \Rightarrow m=\tan \theta \\\ & \Rightarrow m=\tan {{45}^{\circ }} \\\ & \Rightarrow m=1 \\\ \end{aligned}$$ Hence, the slope of this required line is 1. ii) A line which makes an angle of $${{30}^{\circ }}$$ with the positive direction of y-axis measured anti-clockwise Now, as mentioned in the hint, we can use the information as follows As given in the hint, if a line makes some angle the y-axis measured anti-clockwise and if $${{90}^{\circ }}$$ is added to that angle , then the angle that we get is the angle that the same line makes with the x-axis when measured anti-clockwise, hence, the angle that this line which makes an angle of $${{30}^{\circ }}$$ with the positive direction of y-axis measured anti-clockwise, with the x-axis measured anti-clockwise is $$\begin{aligned} & \theta ={{30}^{\circ }}+{{90}^{\circ }} \\\ & \theta ={{120}^{\circ }} \\\ \end{aligned}$$ Now, we can find the slope of this line as follows $$\begin{aligned} & \Rightarrow m=\tan \theta \\\ & \Rightarrow m=\tan {{120}^{\circ }} \\\ & \Rightarrow m=\tan ({{180}^{\circ }}-{{60}^{\circ }}) \\\ & \left[ \tan ({{180}^{\circ }}-x)=-\tan (x) \right] \\\ & \Rightarrow m=-\tan {{60}^{\circ }} \\\ & \Rightarrow m=-\sqrt{3} \\\ \end{aligned}$$ Hence, the slope of this required line is $$-\sqrt{3}$$. Note: The students can make an error if they don’t know about the formulae that are given in the hint as without knowing them one can never get to the correct answer. Also, knowing these following important relations is also very essential which are $$\Rightarrow \tan (\pi -x)=-\tan (x)$$ Also, in this question, it is important to be extra careful while doing the calculations as it might be possible that the students end up getting a wrong result due to some calculation mistakes.