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

Asked by Abhisek | 1 year ago |  126

#### 1 Answer

##### Solution :-

Consider m1 as the slope of the required line

It can be written as

$$y=\dfrac{1}{2}x-\dfrac{3}{2}$$ which is of the form y = mx + c

So the slope of the given line m2 = $$\dfrac{1}{2}$$

We know that the angle between the required line and line x – 2y = 3 is 45°

If θ is the acute angle between lines l1 and l2 with slopes m1 and m2

Substitute the values,

$$1=|\dfrac{\dfrac{1}{2}-m_1}{1+\dfrac{m_1}{2}}|$$

Take LCM,

$$1=|(\dfrac{\dfrac{1-2m_1}{2}}{\dfrac{2+m_1}{2}})|$$

$$1=|\dfrac{1-2m_1}{2+m_1}|$$

$$1=\pm|\dfrac{1-2m_1}{2+m_1}|$$

$$1=|\dfrac{1-2m_1}{2+m_1}|$$ or

$$1=-(\dfrac{1-2m_1}{2+m_1})$$

= 2+m1=1-2m1 or 2+m1 = -1+2m1

$$m_1=-\dfrac{1}{3}$$ or m1= 3

Case 1: m1 = 3

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

y - 2 = 3(x -3)

Expand bracket,

y - 2 = 3x -9

3x - y = 7

Case 2: m1 = $$- \dfrac{1}{3}$$

The equation of the line passing through (3,2) and having a slope of $$- \dfrac{1}{3}$$ is

$$y-2= - \dfrac{1}{3}(x-3)$$

Cross multiply and expand bracket,

3y - 6 = -x +3

x +3y = 9

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

Answered by Pragya Singh | 1 year ago

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