Q79 / 180 · COMEDK 2025 (10 May Shift 3) · Mathematics
The angle between two lines is $45^{\circ}$ and slope of one line is $\dfrac{1}{4}$ then which is the possible value of the slope of the other line.
- A$\dfrac{5}{4}$
- B$\dfrac{4}{5}$
- C$\dfrac{3}{5}$
- Correct answerD$\dfrac{5}{3}$
Solution
Let the slope of the first line be $m_1 = \dfrac{1}{4}$ and the slope of the second line be $m_2$. The angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by the formula $\tan \theta = \left| \dfrac{m_2 - m_1}{1 + m_1 m_2} \right|$.
Given $\theta = 45^{\circ}$, we have $\tan 45^{\circ} = 1$. Substituting the values, we get $1 = \left| \dfrac{m_2 - \dfrac{1}{4}}{1 + \dfrac{1}{4} m_2} \right|$.
This simplifies to $1 = \left| \dfrac{4m_2 - 1}{4 + m_2} \right|$.
This leads to two cases:
Case 1: $\dfrac{4m_2 - 1}{4 + m_2} = 1 \Rightarrow 4m_2 - 1 = 4 + m_2 \Rightarrow 3m_2 = 5 \Rightarrow m_2 = \dfrac{5}{3}$.
Case 2: $\dfrac{4m_2 - 1}{4 + m_2} = -1 \Rightarrow 4m_2 - 1 = -4 - m_2 \Rightarrow 5m_2 = -3 \Rightarrow m_2 = -\dfrac{3}{5}$.
Comparing the results with the given options, the value $\dfrac{5}{3}$ is present.
Answer: $\dfrac{5}{3}$
Given $\theta = 45^{\circ}$, we have $\tan 45^{\circ} = 1$. Substituting the values, we get $1 = \left| \dfrac{m_2 - \dfrac{1}{4}}{1 + \dfrac{1}{4} m_2} \right|$.
This simplifies to $1 = \left| \dfrac{4m_2 - 1}{4 + m_2} \right|$.
This leads to two cases:
Case 1: $\dfrac{4m_2 - 1}{4 + m_2} = 1 \Rightarrow 4m_2 - 1 = 4 + m_2 \Rightarrow 3m_2 = 5 \Rightarrow m_2 = \dfrac{5}{3}$.
Case 2: $\dfrac{4m_2 - 1}{4 + m_2} = -1 \Rightarrow 4m_2 - 1 = -4 - m_2 \Rightarrow 5m_2 = -3 \Rightarrow m_2 = -\dfrac{3}{5}$.
Comparing the results with the given options, the value $\dfrac{5}{3}$ is present.
Answer: $\dfrac{5}{3}$
Hence, the correct answer is option: 'D'.