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NCERT Exemplar · Q7

Q.Find the equation of lines passing through (1,2)(1,2) and making angle 30∘30^\circ with yy-axis.

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A line making angle 30°30° with the yy-axis has slope ±tan⁡60°=±3\pm \tan 60° = \pm \sqrt{3}, giving two lines through (1,2)(1,2): y−2=3(x−1)y - 2 = \sqrt{3}(x-1) and y−2=−3(x−1)y - 2 = -\sqrt{3}(x-1).

Understanding the Geometry

When we say a line makes an angle with the yy-axis, we need to translate that into something we can work with: the slope. The slope of a line is defined by the angle it makes with the positive xx-axis, not the yy-axis.

Here's the key insight: if a line makes angle θ\theta with the yy-axis, it makes angle (90°−θ)(90° - \theta) with the xx-axis. This comes from the fact that the xx and yy axes are perpendicular to each other.

In our problem, the line makes 30°30° with the yy-axis, so it makes 90°−30°=60°90° - 30° = 60° with the xx-axis.

Note

There are actually two lines through any point making a given angle with the yy-axis — one on each side. They make angles θ\theta and 180°−θ180° - \theta with the positive xx-axis.

Finding Both Lines

  1. Identify the two possible angles with the xx-axis

    Since the line makes 30°30° with the yy-axis, it can make either:

    • 60°60° with the positive xx-axis, or
    • 180°−60°=120°180° - 60° = 120° with the positive xx-axis
  2. Calculate the slopes

    The slope mm of a line making angle α\alpha with the positive xx-axis is m=tan⁡αm = \tan \alpha.

    For α=60°\alpha = 60°:

m1=tan⁡60°=3m_1 = \tan 60° = \sqrt{3}

For α=120°\alpha = 120°:

m2=tan⁡120°=tan⁡(180°−60°)=−tan⁡60°=−3m_2 = \tan 120° = \tan(180° - 60°) = -\tan 60° = -\sqrt{3}

  1. Write equations using point-slope form …

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