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Exercise 9.2 · Q4

Q.Passing through (2,23)(2, 2\sqrt{3}) and inclined with the x-axis at an angle of 75∘75^\circ.

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A line through a point with a known angle of inclination has slope m=tan⁡(angle)m = \tan(\text{angle}); here m=tan⁡75∘=2+3m = \tan 75^\circ = 2 + \sqrt{3}, giving the equation y−23=(2+3)(x−2)y - 2\sqrt{3} = (2 + \sqrt{3})(x - 2).

Why slope from angle of inclination works

When a line makes an angle θ\theta with the positive xx-axis (measured counter-clockwise), the slope of that line is precisely m=tan⁡θm = \tan \theta. This comes from the definition of slope as "rise over run": if you move one unit along the xx-axis, the vertical change is tan⁡θ\tan \theta by the geometry of the right triangle formed.

Once we have the slope and a point the line passes through, the point-slope form gives us the equation immediately.

Step-by-step construction

  1. Find the slope from the angle of inclination.

    The line is inclined at 75∘75^\circ to the xx-axis, so

m=tan⁡75∘m = \tan 75^\circ

We need to evaluate tan⁡75∘\tan 75^\circ. Notice that 75∘=45∘+30∘75^\circ = 45^\circ + 30^\circ, so we can use the tangent addition formula:

tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}

With A=45∘A = 45^\circ and B=30∘B = 30^\circ:

tan⁡75∘=tan⁡45∘+tan⁡30∘1−tan⁡45∘tan⁡30∘=1+131−1⋅13=1+131−13\tan 75^\circ = \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}}

Multiply numerator and denominator by 3\sqrt{3}:

tan⁡75∘=3+13−1\tan 75^\circ = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}

Rationalize by multiplying by 3+13+1\frac{\sqrt{3} + 1}{\sqrt{3} + 1}:

tan⁡75∘=(3+1)2(3−1)(3+1)=3+23+13−1=4+232=2+3\tan 75^\circ = \frac{(\sqrt{3} + 1)^2}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{3 + 2\sqrt{3} + 1}{3 - 1} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}

m=tan⁡75∘=2+3m = \tan 75^\circ = 2 + \sqrt{3}

  1. Apply the point-slope form. …

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