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NCERT Exemplar · Q29

Q.The equations of the lines which pass through the point (3,−2)(3,-2) and are inclined at 60∘60^\circ to the line 3x+y=1\sqrt{3}x+y=1 is
(A) y+2=0, 3x−y−2−33=0y+2=0,\ \sqrt{3}x-y-2-3\sqrt{3}=0
(B) x−2=0, 3x−y+2+33=0x-2=0,\ \sqrt{3}x-y+2+3\sqrt{3}=0
(C) 3x−y−2−33=0\sqrt{3}x-y-2-3\sqrt{3}=0
(D) None of these

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The key idea is to use the angle between two lines formula: if a line makes an angle θ\theta with a given line of slope m1m_1, then the slope m2m_2 of the required line satisfies tan⁡θ=∣m2−m11+m1m2∣\tan\theta = \left|\frac{m_2 - m_1}{1 + m_1 m_2}\right|. Here, the given line has slope −3-\sqrt{3}, and the required lines have slopes 00 and 3\sqrt{3}, leading to the two equations y+2=0y+2=0 and 3x−y−2−33=0\sqrt{3}x - y - 2 - 3\sqrt{3}=0. The correct option is (A).


Concept and Intuition

When a problem says "inclined at 60∘60^\circ to a given line," it means the acute angle between the two lines is 60∘60^\circ. The slope of the given line is easy to find. Then, for any unknown line through (3,−2)(3,-2), we set up the angle condition using the tangent formula. This gives two possible slopes — one for each direction the line could be rotated to make a 60∘60^\circ angle. Once we have the slopes, we write the equations using the point-slope form.


Step-by-Step Solution

1. Find the slope of the given line.

The line is 3x+y=1\sqrt{3}x + y = 1. Rewrite in slope-intercept form:

y=−3x+1y = -\sqrt{3}x + 1

So its slope is m1=−3m_1 = -\sqrt{3}.

2. Recall the angle-between-lines formula.

If two lines have slopes m1m_1 and m2m_2, and the acute angle between them is θ\theta, then

tan⁡θ=∣m2−m11+m1m2∣\tan\theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|

Here θ=60∘\theta = 60^\circ, so tan⁡60∘=3\tan 60^\circ = \sqrt{3}.

3. Set up the equation for the unknown slope mm.

Let mm be the slope of a line through (3,−2)(3,-2) that makes 60∘60^\circ with the given line. Then

3=∣m−(−3)1+(−3)m∣=∣m+31−3 m∣\sqrt{3} = \left| \frac{m - (-\sqrt{3})}{1 + (-\sqrt{3}) m} \right| = \left| \frac{m + \sqrt{3}}{1 - \sqrt{3}\, m} \right|

4. Remove the absolute value — two cases.

Case 1:

m+31−3 m=3\frac{m + \sqrt{3}}{1 - \sqrt{3}\, m} = \sqrt{3}

Case 2:

m+31−3 m=−3\frac{m + \sqrt{3}}{1 - \sqrt{3}\, m} = -\sqrt{3}

5. Solve Case 1.

m+3=3(1−3 m)=3−3mm + \sqrt{3} = \sqrt{3}(1 - \sqrt{3}\, m) = \sqrt{3} - 3m

Bring terms together:

m+3m=3−3m + 3m = \sqrt{3} - \sqrt{3}

4m=04m = 0

m=0m = 0

6. Solve Case 2.

m+3=−3(1−3 m)=−3+3mm + \sqrt{3} = -\sqrt{3}(1 - \sqrt{3}\, m) = -\sqrt{3} + 3m

Bring terms:

m−3m=−3−3m - 3m = -\sqrt{3} - \sqrt{3}

−2m=−23-2m = -2\sqrt{3}

m=3m = \sqrt{3}

So the two possible slopes are 00 and 3\sqrt{3}. …

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