Question

سوال اضافی
(۱۶) معادله خطی بنویسید که با جهت مثبت محور x ها زاویه \\(60^{\circ}\\) بسازد و محور عرض ها را به عرض 1 قطع کند؟

Explanation:

🆕 New Concept Discovered: Slope and Angle of Inclination
The slope of a line is the tangent of its angle with the positive x-axis.

Step 1: Find the slope of the line

The slope \( m \) of a line that makes an angle \( \theta \) with the positive direction of the x-axis is given by:

$$ m = \tan(\theta) $$

Given \( \theta = 60^\circ \):

$$ m = \tan(60^\circ) = \sqrt{3} $$

Step 2: Identify the y-intercept

The problem states that the line intersects the y-axis (محور عرض‌ها) at a y-value of \( 1 \).

This means the y-intercept \( b \) is:

$$ b = 1 $$

Step 3: Write the equation of the line

Using the slope-intercept form of a linear equation:

$$ y = mx + b $$

Substitute \( m = \sqrt{3} \) and \( b = 1 \):

$$ y = \sqrt{3}x + 1 $$

Answer:

$$ y = \sqrt{3}x + 1 $$