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Miscellaneous Exercise · Q10

Q.Find the equation of the lines through the point (3,2)(3, 2) which make an angle of 45∘45^\circ with the line x−2y=3x - 2y = 3.

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We use the slope formula for the angle between two lines. The slopes of the required lines are m=3m = 3 and m=−13m = -\frac{1}{3}, giving the equations 3x−y=73x - y = 7 and x+3y=9x + 3y = 9.


The key idea: when a line makes a given angle with another line, the slopes are related by the tangent of the angle between them. Here, the angle is 45∘45^\circ, so tan⁡45∘=1\tan 45^\circ = 1. That gives a clean equation linking the unknown slope mm to the known slope of the given line.

First, find the slope of the given line x−2y=3x - 2y = 3. Rewrite it as y=12x−32y = \frac{1}{2}x - \frac{3}{2}, so its slope is m1=12m_1 = \frac{1}{2}.

Now, if a line with slope mm makes an angle θ\theta with a line of slope m1m_1, then:

tan⁡θ=∣m−m11+mm1∣\tan \theta = \left| \frac{m - m_1}{1 + m m_1} \right|

Here θ=45∘\theta = 45^\circ, so tan⁡45∘=1\tan 45^\circ = 1. Therefore:

∣m−121+m⋅12∣=1\left| \frac{m - \frac{1}{2}}{1 + m \cdot \frac{1}{2}} \right| = 1

This absolute value equation gives two cases — one for the positive, one for the negative.


Step-by-step solution:

  1. Set up the equation without the absolute value.

m−121+m2=±1\frac{m - \frac{1}{2}}{1 + \frac{m}{2}} = \pm 1

  1. Case 1: positive sign.

m−121+m2=1\frac{m - \frac{1}{2}}{1 + \frac{m}{2}} = 1

Multiply numerator and denominator:

m−12=1+m2m - \frac{1}{2} = 1 + \frac{m}{2}

Multiply through by 2:

2m−1=2+m2m - 1 = 2 + m

So m=3m = 3.

  1. Case 2: negative sign.

m−121+m2=−1\frac{m - \frac{1}{2}}{1 + \frac{m}{2}} = -1

Multiply:

m−12=−1−m2m - \frac{1}{2} = -1 - \frac{m}{2}

Multiply by 2:

2m−1=−2−m2m - 1 = -2 - m

So 3m=−13m = -1, giving m=−13m = -\frac{1}{3}.

  1. Find the equations through (3,2)(3, 2). For m=3m = 3:

y−2=3(x−3)⇒y=3x−7y - 2 = 3(x - 3) \quad\Rightarrow\quad y = 3x - 7

Or 3x−y=73x - y = 7. …

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