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Question: The angle between the lines \(kx + y + 9 = 0,y - 3x = 0\) is \({45^ \circ }\), then the value of k i...

The angle between the lines kx+y+9=0,y−3x=0kx + y + 9 = 0,y - 3x = 0 is 45∘{45^ \circ }, then the value of k is
A.2 only
B.2 or −12\dfrac{{ - 1}}{2}
C.– 2 only
D.– 2 and −12\dfrac{{ - 1}}{2}

Explanation

Solution

The angle between two lines is given by the formula .tan⁡θ=∣m2−m11+m1m2∣\tan \theta = \left| {\dfrac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}}} \right|..We just need to calculate the the slope m1{m_1}and m2{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=0kx + y + 9 = 0,y - 3x = 0
We can write an equation of the line in the form of y=mx+cy = 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−9y = - kx - 9
From this we come to know that the slope of the first line m1=−k{m_1} = - k
Now let's write the second equation in the form of y=mx+cy = mx + c
y=3xy = 3x
From this we get that the slope of the second line, m2=3{m_2} = 3
Step 2:
Now the angle between two lines is given by tan⁡θ=∣m2−m11+m1m2∣\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∘{45^ \circ }
Therefore , we get
⇒tan⁡45∘=∣3−(−k)1+(3)(−k)∣\Rightarrow \tan {45^ \circ } = \left| {\dfrac{{3 - ( - k)}}{{1 + (3)( - k)}}} \right|
We now that the value of tan⁡45∘=1\tan {45^ \circ } = 1
⇒±1=3+k1−3k\Rightarrow \pm 1 = \dfrac{{3 + k}}{{1 - 3k}}
Cross multiplying we get
⇒±(1−3k)=3+k\Rightarrow \pm (1 - 3k) = 3 + k
Now let's split it into two cases
Case 1
⇒1−3=k+3k ⇒−2=4k ⇒k=−24=−12  \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 −12\dfrac{{ - 1}}{2}
Case 2 :

⇒−(1−3k)=3+k ⇒−1+3k=3+k ⇒−1−3=k−3k ⇒−4=−2k ⇒k=−4−2=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 −12\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⁡θ=±1m\tan \theta = \dfrac{{ \pm 1}}{m} where ‘m’ is the slope of the other straight line.
If the two lines are a1x+b1y+c1{a_1}x + {b_1}y + {c_1} = 0 and a2x+b2y+c2{a_2}x + {b_2}y + {c_2} = 0, then the formula becomes tan⁡θ=∣a1b2−b1a2a1a2+b1b2∣\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.