Understanding the Problem: Line Through Origin and Angle

The problem asks us to analyze a straight line that passes through the origin \((0,0)\) and makes an angle of \(75^\circ\) with the positive direction of the x-axis. We need to verify two statements about this line.

Finding the Equation of the Line

A line passing through the origin can be represented by the equation \(y = mx\), where \(m\) is the slope of the line. The slope \(m\) is related to the angle \(\theta\) the line makes with the positive x-axis by the formula \(m = \tan(\theta)\).

In this case, the angle is \(\theta = 75^\circ\). So, the slope is \(m = \tan(75^\circ)\).

We can calculate \(\tan(75^\circ)\) using the tangent addition formula: \(\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\).

Let \(A = 45^\circ\) and \(B = 30^\circ\). We know that \(\tan(45^\circ) = 1\) and \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\).

So, \(m = \tan(75^\circ) = \tan(45^\circ + 30^\circ) = \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\frac{\sqrt{3} + 1}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}\).

To simplify the slope, we can multiply the numerator and denominator by the conjugate of the denominator, which is \(\sqrt{3} + 1\):

\(m = \frac{(\sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{(\sqrt{3})^2 + 2\sqrt{3} + 1}{(\sqrt{3})^2 - 1^2} = \frac{3 + 2\sqrt{3} + 1}{3 - 1} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}\).

The equation of the line is \(y = (2 + \sqrt{3})x\).

Analyzing Statement 1: Checking if the Line Passes Through a Specific Point

Statement 1 says the line passes through the point \(\left(1, \frac{1}{2−\sqrt{3}}\right)\). To check this, we substitute the coordinates of the point (\(x=1\), \(y=\frac{1}{2−\sqrt{3}}\)) into the equation of the line \(y = (2 + \sqrt{3})x\).

LHS: \(y = \frac{1}{2−\sqrt{3}}\)

To simplify the LHS, we multiply the numerator and denominator by the conjugate of the denominator, which is \(2 + \sqrt{3}\):

\(y = \frac{1}{2−\sqrt{3}} \cdot \frac{2 + \sqrt{3}}{2 + \sqrt{3}} = \frac{2 + \sqrt{3}}{2^2 - (\sqrt{3})^2} = \frac{2 + \sqrt{3}}{4 - 3} = \frac{2 + \sqrt{3}}{1} = 2 + \sqrt{3}\).

RHS: \((2 + \sqrt{3})x = (2 + \sqrt{3})(1) = 2 + \sqrt{3}\).

Since LHS = RHS (\(2 + \sqrt{3} = 2 + \sqrt{3}\)), the point \(\left(1, \frac{1}{2−\sqrt{3}}\right)\) lies on the line.

Thus, Statement 1 is correct.

Analyzing Statement 2: Determining the Quadrants the Line Lies In

Statement 2 says the line entirely lies in the first and third quadrants.

A line passing through the origin (\(0,0\)) will lie in the first and third quadrants if its slope is positive. It will lie in the second and fourth quadrants if its slope is negative. It will lie only on axes if the angle is \(0^\circ\) or \(90^\circ\).

The angle the line makes with the positive x-axis is \(75^\circ\). Since \(0^\circ < 75^\circ < 90^\circ\), the angle is in the first quadrant. The tangent of an angle in the first quadrant is positive.

We calculated the slope \(m = \tan(75^\circ) = 2 + \sqrt{3}\). Since \(2 + \sqrt{3}\) is a positive value, the slope of the line is positive.

A line with a positive slope passing through the origin rises from the third quadrant through the origin into the first quadrant.

Therefore, the line entirely lies in the first and third quadrants.

Thus, Statement 2 is correct.

Conclusion

Both Statement 1 and Statement 2 are correct.

Revision Table: Line Properties

Additional Information: Coordinate Geometry Concepts

Coordinate geometry is a branch of mathematics where geometric problems are solved using a coordinate system. The Cartesian coordinate system is commonly used, consisting of two perpendicular axes (x and y) intersecting at the origin (0,0).

• Slope of a Line: The slope \(m\) represents the steepness and direction of a line. It is the ratio of the vertical change to the horizontal change between any two distinct points on the line.
• Equation of a Line: There are various forms to represent a line's equation, such as the slope-intercept form \(y = mx + c\) (where \(c\) is the y-intercept) or the point-slope form \(y - y_1 = m(x - x_1)\). For a line through the origin, the y-intercept \(c\) is 0, simplifying to \(y = mx\).
• Quadrants: The coordinate plane is divided into four quadrants by the x and y axes.
  • Quadrant I: \(x > 0, y > 0\)
  • Quadrant II: \(x < 0, y > 0\)
  • Quadrant III: \(x < 0, y < 0\)
  • Quadrant IV: \(x > 0, y < 0\)
• Angle with x-axis: The angle \(\theta\) is measured counterclockwise from the positive x-axis to the line. The slope \(m\) is equal to \(\tan(\theta)\).