Question 15 of 70intermediate🔧 ApplyVery Short2 marks

In Exercises 1 to 8, find the equation of the line which satisfy the given conditions: Passing through (2,23)(2, 2\sqrt{3}) and inclined with the xx-axis at an angle of 75∘75^\circ.

Correct Answer

The equation of the line is (2+3)x−y−4=0(2 + \sqrt{3})x - y - 4 = 0.

Finding the slope: Since the line is inclined at 75°75° with the xx-axis, the slope is m=tan⁡75°m = \tan 75°.

m=tan⁡(45°+30°)=tan⁡45°+tan⁡30°1−tan⁡45°⋅tan⁡30°=1+131−13=3+13−1=2+3m = \tan(45° + 30°) = \frac{\tan 45° + \tan 30°}{1 - \tan 45° \cdot \tan 30°} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} = 2 + \sqrt{3}

Using point-slope form: Using the point (2,23)(2, 2\sqrt{3}) and slope m=2+3m = 2 + \sqrt{3}:

y−23=(2+3)(x−2)y - 2\sqrt{3} = (2 + \sqrt{3})(x - 2)

y=(2+3)x−4y = (2 + \sqrt{3})x - 4

Therefore, the required equation is (2+3)x−y−4=0(2 + \sqrt{3})x - y - 4 = 0.

Exercise: EXERCISE 9.2 | Q: 4 | (Page 13)
For More Understanding

Explanation

This question tests two concepts from the chapter: (1) finding slope from angle of inclination using m=tan⁡θm = \tan\theta, and (2) using the point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1) to find the equation of a line. The context shows in Example 1(d) that slope is found from inclination angle, and Example 16 demonstrates the point-slope form. The calculation of tan⁡75°\tan 75° using compound angle formula tan⁡(A+B)\tan(A+B) is a standard trigonometric result that students should know.

Solution Steps

  1. Step 1: Calculate slope m=tan⁡75°=tan⁡(45°+30°)=2+3m = \tan 75° = \tan(45° + 30°) = 2 + \sqrt{3}

  2. Step 2: Apply point-slope form: y−23=(2+3)(x−2)y - 2\sqrt{3} = (2 + \sqrt{3})(x - 2)

  3. Step 3: Simplify to get (2+3)x−y−4=0(2 + \sqrt{3})x - y - 4 = 0

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