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NCERT Exemplar · Q3

Q.Find the angle between the lines y=(2−3)(x+5)y=(2-\sqrt{3})(x+5) and y=(2+3)(x−7)y=(2+\sqrt{3})(x-7).

Rajasthan RbseShort· 3mImportance★★★★★est
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✓ Free question

The angle between two lines is found using their slopes m1m_1 and m2m_2 with the formula tan⁡θ=∣m1−m21+m1m2∣\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|. For the given lines, the slopes are 2−32-\sqrt{3} and 2+32+\sqrt{3}, leading to an angle of 60∘\boxed{60^\circ}.

When we talk about the angle between two lines, we are typically referring to the acute angle formed by their intersection. The key to finding this angle lies in understanding what the slope of a line represents.

The slope mm of a line is defined as the tangent of the angle α\alpha that the line makes with the positive direction of the x-axis. That is, m=tan⁡αm = \tan \alpha. This angle α\alpha is measured counter-clockwise from the positive x-axis to the line.

If we have two lines with slopes m1m_1 and m2m_2, making angles α1\alpha_1 and α2\alpha_2 respectively with the positive x-axis, then the angle θ\theta between these two lines can be found using the tangent subtraction formula.

The angle θ\theta between two lines with slopes m1m_1 and m2m_2 is given by:

tan⁡θ=∣m1−m21+m1m2∣\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|

The absolute value ensures that we find the acute angle between the lines. If 1+m1m2=01 + m_1 m_2 = 0, it means m1m2=−1m_1 m_2 = -1, which implies the lines are perpendicular, and the angle is 90∘90^\circ.

Let's apply this concept to the given problem.

  1. Identify the slopes of the lines.

    The equations of the lines are given in the form y=mx+cy = mx + c, where mm is the slope and cc is the y-intercept.

    For the first line, y=(2−3)(x+5)y = (2-\sqrt{3})(x+5), the slope m1m_1 is 2−32-\sqrt{3}.

    For the second line, y=(2+3)(x−7)y = (2+\sqrt{3})(x-7), the slope m2m_2 is 2+32+\sqrt{3}.

  2. Substitute the slopes into the angle formula.

    We use the formula tan⁡θ=∣m1−m21+m1m2∣\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|.

    Let's calculate the numerator and denominator separately.

    • Numerator: m1−m2m_1 - m_2

m1−m2=(2−3)−(2+3)m_1 - m_2 = (2-\sqrt{3}) - (2+\sqrt{3})

m1−m2=2−3−2−3m_1 - m_2 = 2 - \sqrt{3} - 2 - \sqrt{3}

m1−m2=−23m_1 - m_2 = -2\sqrt{3}

*   **Denominator:** $1 + m_1 m_2$

1+m1m2=1+(2−3)(2+3)1 + m_1 m_2 = 1 + (2-\sqrt{3})(2+\sqrt{3})

    Notice that the product $(2-\sqrt{3})(2+\sqrt{3})$ is in the form $(a-b)(a+b)$, which simplifies to $a^2 - b^2$.
    Here, $a=2$ and $b=\sqrt{3}$.

(2−3)(2+3)=22−(3)2(2-\sqrt{3})(2+\sqrt{3}) = 2^2 - (\sqrt{3})^2

=4−3= 4 - 3

=1= 1

    So, the denominator becomes:

1+m1m2=1+11 + m_1 m_2 = 1 + 1

=2= 2

  1. Calculate tan⁡θ\tan \theta. Now, substitute the calculated numerator and denominator back into the formula:

tan⁡θ=∣−232∣\tan \theta = \left| \frac{-2\sqrt{3}}{2} \right|

tan⁡θ=∣−3∣\tan \theta = \left| -\sqrt{3} \right|

tan⁡θ=3\tan \theta = \sqrt{3}

  1. Find the angle θ\theta. We need to find the angle θ\theta whose tangent is 3\sqrt{3}. We know that tan⁡60∘=3\tan 60^\circ = \sqrt{3}. Therefore, θ=60∘\theta = 60^\circ.
✓Final answer

The angle between the lines is 60∘\boxed{60^\circ}.

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