Question:medium
If the line joining of two points \((1,0)\) and \((4,3)\) is rotated about the point \((1,0)\) in counter clockwise direction through an angle \(15^\circ\), then the equation of the line in the new position is
If the line joining of two points \((1,0)\) and \((4,3)\) is rotated about the point \((1,0)\) in counter clockwise direction through an angle \(15^\circ\), then the equation of the line in the new position is
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When a line is rotated by an angle, first find its original inclination, then add or subtract the rotation angle, and finally use the new slope with the fixed point.
Updated On: May 14, 2026
- \(\sqrt{3}x-2y-\sqrt{3}=0\)
- \(\sqrt{3}x-y-\sqrt{3}=0\)
- \(x+y-1=0\)
- \(x+\sqrt{3}y-1=0\)
- \(3x-y-3=0\)
Show Solution
The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
We are given a line segment defined by two points. This line is rotated around one of its endpoints. We need to find the equation of the new line. The key is to find the new slope of the line after rotation. The equation of a line can be found using the point-slope form, \(y - y_1 = m(x - x_1)\), once we have a point on the line (the pivot point) and the new slope.
Step 2: Key Formula or Approach:
1. Find the initial slope (\(m_1\)) of the line passing through (1, 0) and (4, 3). 2. The slope is the tangent of the angle of inclination (\(\theta_1\)), so \(m_1 = \tan(\theta_1)\). 3. The new line is rotated counter-clockwise by an angle \(\alpha = 15^\circ\). The new angle of inclination will be \(\theta_2 = \theta_1 + \alpha\). 4. The new slope will be \(m_2 = \tan(\theta_2) = \tan(\theta_1 + \alpha)\). 5. Use the tangent addition formula: \(\tan(\theta_1 + \alpha) = \frac{\tan(\theta_1) + \tan(\alpha)}{1 - \tan(\theta_1)\tan(\alpha)}\). 6. Use the point-slope form to find the equation of the new line, which passes through (1, 0) with slope \(m_2\).
Step 3: Detailed Explanation:
1. Find the initial slope \(m_1\):
The line passes through A(1, 0) and B(4, 3).
\[ m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3-0}{4-1} = \frac{3}{3} = 1 \] So, the initial slope is \(m_1 = 1\).
2. Find the initial angle of inclination \(\theta_1\):
\(m_1 = \tan(\theta_1) = 1\). This implies \(\theta_1 = 45^\circ\) (or \(\pi/4\)).
3. Find the new angle of inclination \(\theta_2\):
The line is rotated counter-clockwise by \(\alpha = 15^\circ\).
\[ \theta_2 = \theta_1 + \alpha = 45^\circ + 15^\circ = 60^\circ \] 4. Find the new slope \(m_2\):
\[ m_2 = \tan(\theta_2) = \tan(60^\circ) = \sqrt{3} \] 5. Find the equation of the new line:
The new line has slope \(m_2 = \sqrt{3}\) and passes through the point of rotation (1, 0).
Using the point-slope form \(y - y_1 = m_2(x - x_1)\):
\[ y - 0 = \sqrt{3}(x - 1) \] \[ y = \sqrt{3}x - \sqrt{3} \] Rearranging the terms to match the options:
\[ \sqrt{3}x - y - \sqrt{3} = 0 \] Step 4: Final Answer:
The equation of the line in the new position is \(\sqrt{3}x - y - \sqrt{3} = 0\). This corresponds to option (B).
We are given a line segment defined by two points. This line is rotated around one of its endpoints. We need to find the equation of the new line. The key is to find the new slope of the line after rotation. The equation of a line can be found using the point-slope form, \(y - y_1 = m(x - x_1)\), once we have a point on the line (the pivot point) and the new slope.
Step 2: Key Formula or Approach:
1. Find the initial slope (\(m_1\)) of the line passing through (1, 0) and (4, 3). 2. The slope is the tangent of the angle of inclination (\(\theta_1\)), so \(m_1 = \tan(\theta_1)\). 3. The new line is rotated counter-clockwise by an angle \(\alpha = 15^\circ\). The new angle of inclination will be \(\theta_2 = \theta_1 + \alpha\). 4. The new slope will be \(m_2 = \tan(\theta_2) = \tan(\theta_1 + \alpha)\). 5. Use the tangent addition formula: \(\tan(\theta_1 + \alpha) = \frac{\tan(\theta_1) + \tan(\alpha)}{1 - \tan(\theta_1)\tan(\alpha)}\). 6. Use the point-slope form to find the equation of the new line, which passes through (1, 0) with slope \(m_2\).
Step 3: Detailed Explanation:
1. Find the initial slope \(m_1\):
The line passes through A(1, 0) and B(4, 3).
\[ m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3-0}{4-1} = \frac{3}{3} = 1 \] So, the initial slope is \(m_1 = 1\).
2. Find the initial angle of inclination \(\theta_1\):
\(m_1 = \tan(\theta_1) = 1\). This implies \(\theta_1 = 45^\circ\) (or \(\pi/4\)).
3. Find the new angle of inclination \(\theta_2\):
The line is rotated counter-clockwise by \(\alpha = 15^\circ\).
\[ \theta_2 = \theta_1 + \alpha = 45^\circ + 15^\circ = 60^\circ \] 4. Find the new slope \(m_2\):
\[ m_2 = \tan(\theta_2) = \tan(60^\circ) = \sqrt{3} \] 5. Find the equation of the new line:
The new line has slope \(m_2 = \sqrt{3}\) and passes through the point of rotation (1, 0).
Using the point-slope form \(y - y_1 = m_2(x - x_1)\):
\[ y - 0 = \sqrt{3}(x - 1) \] \[ y = \sqrt{3}x - \sqrt{3} \] Rearranging the terms to match the options:
\[ \sqrt{3}x - y - \sqrt{3} = 0 \] Step 4: Final Answer:
The equation of the line in the new position is \(\sqrt{3}x - y - \sqrt{3} = 0\). This corresponds to option (B).
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