Question

The angle between the lines $kx + y + 9 = 0,y - 3x = 0$ is ${45^ \circ }$, then the value of k is
A.2 only
B.2 or $\dfrac{{ - 1}}{2}$
C.– 2 only
D.– 2 and $\dfrac{{ - 1}}{2}$

Answer 1

Hint: The angle between two lines is given by the formula .$\tan \theta = \left| {\dfrac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}}} \right|$..We just need to calculate the the slope ${m_1}$and ${m_2}$ and substitute in the formula to find the value of k

Complete step-by-step answer:
Step 1 : We are the equations of the line to be $kx + y + 9 = 0,y - 3x = 0$
We can write an equation of the line in the form of $y = mx + c$ when the coefficient of x , m is the slope of the line.
So now let's write the first equation in this form
$y = - kx - 9$
From this we come to know that the slope of the first line ${m_1} = - k$
Now let's write the second equation in the form of $y = mx + c$
$y = 3x$
From this we get that the slope of the second line, ${m_2} = 3$
Step 2:
Now the angle between two lines is given by $\tan \theta = \left| {\dfrac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}}} \right|$
And we are given that the angle between these given lines is ${45^ \circ }$
Therefore , we get
$ \Rightarrow \tan {45^ \circ } = \left| {\dfrac{{3 - ( - k)}}{{1 + (3)( - k)}}} \right|$
We now that the value of $\tan {45^ \circ } = 1$
$ \Rightarrow \pm 1 = \dfrac{{3 + k}}{{1 - 3k}}$
Cross multiplying we get
$ \Rightarrow \pm (1 - 3k) = 3 + k$
Now let's split it into two cases
Case 1
$
   \Rightarrow 1 - 3 = k + 3k \\
   \Rightarrow - 2 = 4k \\
   \Rightarrow k = \dfrac{{ - 2}}{4} = \dfrac{{ - 1}}{2} \\
$
In this case we get the value of k to be \[\dfrac{{ - 1}}{2}\]
Case 2 :
\[
   \Rightarrow - (1 - 3k) = 3 + k \\
   \Rightarrow - 1 + 3k = 3 + k \\
   \Rightarrow - 1 - 3 = k - 3k \\
   \Rightarrow - 4 = - 2k \\
   \Rightarrow k = \dfrac{{ - 4}}{{ - 2}} = 2 \\
\]
In this case we get the value of k to be 2
Therefore the value of k is \[\dfrac{{ - 1}}{2}\] and 2
The correct option is B

Note: If one of the line is parallel to y-axis then the angle between two straight lines is given by $\tan \theta = \dfrac{{ \pm 1}}{m}$ where ‘m’ is the slope of the other straight line.
If the two lines are ${a_1}x + {b_1}y + {c_1}$ = 0 and ${a_2}x + {b_2}y + {c_2}$ = 0, then the formula becomes $\tan \theta = \left| {\dfrac{{{a_1}{b_2} - {b_1}{a_2}}}{{{a_1}{a_2} + {b_1}{b_2}}}} \right|$
Generally speaking, the angle between these two lines is assumed to be acute and hence, the value of tan θ is taken to be positive.