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To find the equation of the line that makes an angle of \(120^\circ\) with the X-axis and passes through the point \((2, 3)\), we can proceed as follows:

### Step 1: Find the slope of the line
The slope \(m\) of a line that makes an angle \(\theta\) with the X-axis is given by:
\[
m = \tan(\theta)
\]
Here, \(\theta = 120^\circ\). So,
\[
m = \tan(120^\circ)
\]
Using the tangent of \(120^\circ\):
\[
\tan(120^\circ) = \tan(180^\circ - 60^\circ) = -\tan(60^\circ)
\]
Since \(\tan(60^\circ) = \sqrt{3}\), we have:
\[
m = -\sqrt{3}
\]

### Step 2: Use the point-slope form of the equation of a line
The point-slope form of the equation of a line is:
\[
y - y_1 = m(x - x_1)
\]
where \((x_1, y_1)\) is a point on the line and \(m\) is the slope. Here, \((x_1, y_1) = (2, 3)\) and \(m = -\sqrt{3}\).

Substitute the values:
\[
y - 3 = -\sqrt{3}(x - 2)
\]

### Step 3: Simplify the equation
Expand and simplify the equation:
\[
y - 3 = -\sqrt{3}x + 2\sqrt{3}
\]
\[
y = -\sqrt{3}x + 2\sqrt{3} + 3
\]

Thus, the equation of the line is:
\[
y = -\sqrt{3}x + (2\sqrt{3} + 3)
\]

Let me know if you need further clarifications.

### Related Questions:
1. What is the equation of a line passing through the origin and making an angle of \(45^\circ\) with the X-axis?
2. How do you derive the general form of the equation of a line given its slope and a point?
3. What happens to the slope if the angle made with the X-axis is \(90^\circ\)?
4. How do you find the equation of a line perpendicular to the line obtained here?
5. How can we check if two lines are parallel based on their equations?

### Tip:
Always remember that the slope of a line is negative when the angle with the X-axis is between \(90^\circ\) and \(180^\circ\).

Find the equation of a line making an angle of 120° with X-axis and passing through (2, 3).

Learn how to find the equation of a line that makes an angle of 120° with the X-axis and passes through the point (2, 3). This solution covers key algebraic concepts such as slope-angle relationships and the point-slope form of a line. Ideal for high school students in Grades 9-12.

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