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Exercise 9.2 · Q6

Q.Intersecting the y-axis at a distance of 22 units above the origin and making an angle of 30∘30^\circ with positive direction of the x-axis.

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The line passes through (0,2)(0,2) with slope tan⁡30∘=13\tan 30^\circ = \frac{1}{\sqrt{3}}, so its equation is y=13x+2y = \frac{1}{\sqrt{3}}x + 2.

Why this approach works

Every straight line in the plane is completely determined by two pieces of information. Here we are given:

  • A specific point: the y-intercept is at (0,2)(0,2) — that's "2 units above the origin" along the y-axis.
  • A direction: the line makes 30∘30^\circ with the positive x-axis.

The slope of a line is defined as tan⁡θ\tan \theta, where θ\theta is the angle it makes with the positive x-axis. Once we have the slope mm and a point (x1,y1)(x_1, y_1), the point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1) gives the equation directly. Since the given point is the y-intercept, the equation simplifies beautifully to the slope-intercept form y=mx+cy = mx + c, where cc is the y-intercept.

Step-by-step solution

  1. Identify the given point.

    "Intersecting the y-axis at a distance of 2 units above the origin" means the line crosses the y-axis at y=2y = 2. So the point is (0,2)(0, 2).

  2. Find the slope from the given angle.

    The line makes an angle of 30∘30^\circ with the positive x-axis.

    Slope m=tan⁡30∘=13m = \tan 30^\circ = \frac{1}{\sqrt{3}}.

    Tip

    Remember: tan⁡30∘=13\tan 30^\circ = \frac{1}{\sqrt{3}}, tan⁡45∘=1\tan 45^\circ = 1, tan⁡60∘=3\tan 60^\circ = \sqrt{3}. These three are the most common in exam problems.

  3. Write the equation using the slope-intercept form. …

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