Question

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

• A. \(\sqrt{3}x-2y-\sqrt{3}=0\)
• B. \(\sqrt{3}x-y-\sqrt{3}=0\)
• C. \(x+y-1=0\)
• D. \(x+\sqrt{3}y-1=0\)
• E. \(3x-y-3=0\)

Hint:

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.

Answer 1

The Correct Option is B

Step 1: Find the slope of the original line.
The given line passes through the points \((1,0)\) and \((4,3)\). Its slope is:
\[ m=\frac{3-0}{4-1}=\frac{3}{3}=1 \] So the line makes an angle \(\theta\) with the positive \(x\)-axis such that:
\[ \tan \theta =1 \] Hence,
\[ \theta=45^\circ \]

Step 2: Understand the effect of rotation.
The line is rotated counterclockwise by \(15^\circ\) about the point \((1,0)\). Since rotation changes the inclination of the line, the new angle with the positive \(x\)-axis becomes:
\[ 45^\circ+15^\circ=60^\circ \]

Step 3: Find the slope of the rotated line.
The slope of a line making angle \(60^\circ\) with the positive \(x\)-axis is:
\[ m'=\tan 60^\circ=\sqrt{3} \]

Step 4: Use the point through which the new line passes.
Because the rotation is about the point \((1,0)\), that point remains fixed. So the new line passes through:
\[ (1,0) \]

Step 5: Write the point-slope form of the new line.
Using slope \(m'=\sqrt{3}\) and point \((1,0)\):
\[ y-0=\sqrt{3}(x-1) \] \[ y=\sqrt{3}(x-1) \]

Step 6: Convert into standard form.
Expanding:
\[ y=\sqrt{3}x-\sqrt{3} \] Bringing all terms to one side:
\[ \sqrt{3}x-y-\sqrt{3}=0 \]

Step 7: Match with the options.
The equation \(\sqrt{3}x-y-\sqrt{3}=0\) matches option \((2)\). Therefore, the correct answer is:
\[ \boxed{\sqrt{3}x-y-\sqrt{3}=0} \]