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JEE Main 2020
Straight Lines
Straight Lines and Pair of Straight Lines
Medium

Question

A square, of each side 2, lies above the x-axis and has one vertex at the origin. If one of the sides passing through the origin makes an angle 30 o with the positive direction of the x-axis, then the sum of the x-coordinates of the vertices of the square is :

Options

Solution

Key Concepts and Formulas

  • Polar Coordinates: A point PP at distance rr from the origin, with the line segment OPOP making an angle θ\theta with the positive x-axis, has coordinates (rcosθ,rsinθ)(r \cos \theta, r \sin \theta).
  • Properties of a Square: All sides are equal in length, and all interior angles are 90 degrees.
  • Trigonometric Values: cos(30)=32\cos(30^\circ) = \frac{\sqrt{3}}{2}, sin(30)=12\sin(30^\circ) = \frac{1}{2}, cos(120)=12\cos(120^\circ) = -\frac{1}{2}, sin(120)=32\sin(120^\circ) = \frac{\sqrt{3}}{2}

Step-by-Step Solution

Step 1: Define the Vertices and Angles

Let the square be OABCOABC, where OO is the origin (0,0)(0,0). The side length of the square is s=2s=2. OAOA makes an angle of 3030^\circ with the positive x-axis. Since OABCOABC is a square, OCOC makes an angle of 30+90=12030^\circ + 90^\circ = 120^\circ with the positive x-axis.

Step 2: Calculate the Coordinates of Vertex A

Vertex AA is at a distance of 22 from the origin and makes an angle of 3030^\circ with the x-axis. Using polar coordinates: A=(2cos30,2sin30)=(232,212)=(3,1)A = (2 \cos 30^\circ, 2 \sin 30^\circ) = \left(2 \cdot \frac{\sqrt{3}}{2}, 2 \cdot \frac{1}{2}\right) = (\sqrt{3}, 1)

Step 3: Calculate the Coordinates of Vertex C

Vertex CC is at a distance of 22 from the origin and makes an angle of 120120^\circ with the x-axis. Using polar coordinates: C=(2cos120,2sin120)=(2(12),232)=(1,3)C = (2 \cos 120^\circ, 2 \sin 120^\circ) = \left(2 \cdot \left(-\frac{1}{2}\right), 2 \cdot \frac{\sqrt{3}}{2}\right) = (-1, \sqrt{3})

Step 4: Calculate the Coordinates of Vertex B

Vertex BB can be found by recognizing that OB=OA+OC\vec{OB} = \vec{OA} + \vec{OC}. Thus, the coordinates of BB are the sum of the coordinates of AA and CC. B=(3+(1),1+3)=(31,1+3)B = (\sqrt{3} + (-1), 1 + \sqrt{3}) = (\sqrt{3} - 1, 1 + \sqrt{3})

Step 5: Find the Sum of the x-coordinates

The x-coordinates of the vertices are: xO=0x_O = 0 xA=3x_A = \sqrt{3} xB=31x_B = \sqrt{3} - 1 xC=1x_C = -1

The sum of the x-coordinates is: 0+3+(31)+(1)=2320 + \sqrt{3} + (\sqrt{3} - 1) + (-1) = 2\sqrt{3} - 2

Common Mistakes & Tips

  • Angle Calculation: Be careful when calculating the angles for each vertex. Ensure you are adding or subtracting 9090^\circ correctly based on the orientation of the square.
  • Vector Addition: Using vector addition to find the fourth vertex is generally simpler than trying to find the angle it makes with the x-axis.
  • Diagram: Drawing a clear diagram can help visualize the problem and avoid errors in angle calculations.

Summary

The problem involves finding the coordinates of the vertices of a square given its side length and the angle one of its sides makes with the x-axis. We used polar coordinates to find the coordinates of two vertices and vector addition to find the coordinates of the fourth. Finally, we summed the x-coordinates of all four vertices. The sum of the x-coordinates of the vertices of the square is 2322\sqrt{3} - 2.

Final Answer

The final answer is \boxed{2\sqrt 3 - 2}, which corresponds to option (B).

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