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Question: What is the acute angle between the lines \[y = 3x + 2\] and \[y = 4x + 9\]? A. \[{4.4^ \circ }\] ...

What is the acute angle between the lines y=3x+2y = 3x + 2 and y=4x+9y = 4x + 9?
A. 4.4{4.4^ \circ }
B. 28.3{28.3^ \circ }
C. 5.2{5.2^ \circ }
D. 18.6{18.6^ \circ }
E. none of these

Explanation

Solution

Here, we are given two lines and we need to find the acute angle between them. First, we will find the slope of the line and since they are in slope intercept form (i.e. y=mx+cy = mx + c ) where m is the slope of the line. Next, after getting slope of both the lines we will use the formula to find the angle between them i.e. tanθ=m2m11+m1m2\tan \theta = \dfrac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}} . Thus, solving this, we will get the final output.

Complete step by step answer:
Given that, there are two lines of the equations: y=3x+2y = 3x + 2 and y=4x+9y = 4x + 9. We know that an angle is called an acute angled triangle if all the angles of the triangle are less than 90 degrees. As we can clearly identify the slope of the equation from the given equations as they are in the form of slope intercept i.e. y=mx+cy = mx + c where m is the slope of the equation, c is the constant and points (x, y) are the coordinates.So,
Slope of the equation y=3x+2y = 3x + 2 will be
m1=3{m_1} = 3
tanθ1=3\Rightarrow \tan {\theta _1} = 3
θ1=tan1(3)\Rightarrow {\theta _1} = {\tan ^{ - 1}}(3)
θ1=71.57\Rightarrow {\theta _1} = {71.57^ \circ }

And, slope of the equation y=4x+9y = 4x + 9 will be
m2=4{m_2} = 4
tanθ2=4\Rightarrow \tan {\theta _2} = 4
θ2=tan1(4)\Rightarrow {\theta _2} = {\tan ^{ - 1}}(4)
θ2=75.96\Rightarrow {\theta _2} = {75.96^ \circ }
Thus, the angles between the lines is
θ2θ1=75.9671.57{\theta _2} - {\theta _1} = {75.96^ \circ } - {71.57^ \circ }
θ2θ1=4.4\therefore {\theta _2} - {\theta _1} = {4.4^ \circ }

Another Method: Angles between the two lines is:
tanθ=m2m11+m1m2\tan \theta = \dfrac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}}
θ=tan1(m2m11+m1m2)\Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}}} \right)
Substituting the values of m1 and m2, we will get,
θ=tan1(431+(3)(4))\Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{{4 - 3}}{{1 + (3)(4)}}} \right)
On evaluating this, we will get,
θ=tan1(11+12)\Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{1}{{1 + 12}}} \right)
θ=tan1(113)\Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{1}{{13}}} \right)
θ=4.4\therefore \theta = {4.4^ \circ }

Hence, the acute angles between the lines is 4.4{4.4^ \circ }.

Note: We know that, we slopes are equal then, it means that lines are parallel and angles between them is θ=0\theta = {0^ \circ }. And if the product of the slopes is equal to -1, it means that lines are perpendicular and angles between them is θ=90\theta = {90^ \circ }. The value of tanθ\tan \theta will always be positive.