The obtuse angle between lines \(2y = x + 1\) and \(y = 3x + 2\) is:
\(3\pi/4\)
\(5\pi/6\)
\(4\pi/3\)
\(2\pi/3\)
Step 1: Write both equations in slope-intercept form \(y = mx + c\).
For \(2y = x + 1 \Rightarrow y = \frac{1}{2}x + \frac{1}{2}\). Hence, slope \(m_1 = \frac{1}{2}\).
For \(y = 3x + 2\), slope \(m_2 = 3\). Step 2: The acute angle \(\theta\) between two lines with slopes \(m_1\) and \(m_2\) is given by: \[ \tan\theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right| \] Substitute values: \[ \tan\theta = \left| \frac{3 - \frac{1}{2}}{1 + 3 \times \frac{1}{2}} \right| = \left| \frac{\frac{5}{2}}{\frac{5}{2}} \right| = 1 \] \[ \Rightarrow \theta = \frac{\pi}{4} \] Step 3: The question asks for the obtuse angle between the lines. The obtuse angle is: \[ \pi - \frac{\pi}{4} = \frac{3\pi}{4} \] Final Answer: \[ \boxed{\text{Obtuse angle} = \frac{3\pi}{4}} \] Correct Option: (a)
What is the general solution of the equation \( \cot\theta + \tan\theta = 2 \)?
Let the line $\frac{x}{4} + \frac{y}{2} = 1$ meet the x-axis and y-axis at A and B, respectively. M is the midpoint of side AB, and M' is the image of the point M across the line $x + y = 1$. Let the point P lie on the line $x + y = 1$ such that $\Delta ABP$ is an isosceles triangle with $AP = BP$. Then the distance between M' and P is:
Let \( F_1, F_2 \) \(\text{ be the foci of the hyperbola}\) \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, a > 0, \, b > 0, \] and let \( O \) be the origin. Let \( M \) be an arbitrary point on curve \( C \) and above the X-axis and \( H \) be a point on \( MF_1 \) such that \( MF_2 \perp F_1 F_2, \, M F_1 \perp OH, \, |OH| = \lambda |O F_2| \) with \( \lambda \in (2/5, 3/5) \), then the range of the eccentricity \( e \) is in: