# Find the equation of the lines through the point (3, 2) which make an angle of 45° with the line x - 2y = 3

**Solution:**

Let the slope of the required line be m\(_1\)

The given line can be represented as y = 1/2x - 3/2, which is of the form y = mx + c

Slope of the given line = m\(_2\) = 1/2

It is given that the angle between the required line and line x - 2y = 3 is 45°.

We know that if θ is the acute angle between lines l_{1} and l_{2} with the slopes m_{1} and m_{2} respectively,

Then,

tanθ = |(m\(_2\) - m\(_1\))/(1 + m\(_1\)m\(_2\))|

tan 45° = |(m\(_2\) - m\(_1\))/(1 + m\(_1\)m\(_2\))|

1 = |1/2 - m\(_1\)/(1 + m\(_1\)/2)|

1 = |(1 - 2m\(_1\)/2)/(2 + m\(_1\))/2)|

1 = ± (1 - 2m\(_1\))/(2 + m\(_1\))

1 = (1 - 2m\(_1\))/(2 + m\(_1\)) or 1 = - (1 - 2m\(_1\))/(2 + m\(_1\))

⇒ 2 + m\(_1\) = 1 - 2m\(_1\) or ⇒ 2 + m\(_1\) = - 1 + 2m\(_1\)

m\(_1\) = - 1/3 or m\(_1\) = 3

__Case I__**:** m\(_1\) = 3

The equation of the line passing through (3, 2) and having a slope of 3 is:

y - 2 = 3(x - 3)

y - 2 = 3x - 9

3x - y = 7

**Case II:** m\(_1\) = - 1/3

The equation of the line passing through (3,2) and having a slope of - 1/3 is

y - 2 = - 1/3 (x - 2)

3y - 6 = - x + 3

x + 3y = 9

Thus, the equations of the line are 3x - y = 7 and x + 3y = 9

NCERT Solutions Class 11 Maths Chapter 10 Exercise ME Question 11

## Find the equation of the lines through the point (3, 2) which make an angle of 45° with the line x - 2y = 3

**Summary:**

The equations of the lines through the point (3, 2) which make an angle of 45° with the line x - 2y = 3 are 3x - y = 7 and x + 3y = 9

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