Math, asked by irfanafathima631, 4 months ago

find the angle between the line y-✓3-5=0 and ✓3-X+6=0​

Answers

Answered by MODAZEEM
1

Answer:

y - \sqrt{3} x - 5 = 0 ---- (1)y−

3

x−5=0−−−−(1)

y = \sqrt{3} x + 5y=

3

x+5

It is in the form of y1 = m1x1 + c.

m1 = \sqrt{3} m1=

3

\sqrt{3} y - x + 6 = 0

3

y−x+6=0 ------ (2)

\sqrt{3} y = x - 6

3

y=x−6

y = \frac{1}{ \sqrt{3} } x - \frac{6}{ \sqrt{3} } y=

3

1

x−

3

6

y = \frac{1}{ \sqrt{3} } x - 2 \sqrt{3} y=

3

1

x−2

3

It is in the form of y2 = m2x2 + c.

m2 = \frac{1}{ \sqrt{3} } m2=

3

1

Now,

Note: Here I am writing ∅ as A because it is difficult for me to write ∅ always.

Let A be the angle between the lines.Then

tan A = \frac{m1 - m2}{1 + m1m2} tanA=

1+m1m2

m1−m2

= \frac{ \sqrt{3} - \frac{1}{ \sqrt{3} } }{1 + \sqrt{3} * \frac{1}{ \sqrt{3} } } =

1+

3

3

1

3

3

1

= \frac{ \sqrt{3} - \frac{1}{ \sqrt{3} } }{1 + 1} =

1+1

3

3

1

= \frac{2}{ \frac{ \sqrt{3} }{2} } =

2

3

2

= \frac{2}{2 \sqrt{3} } =

2

3

2

= \frac{1}{ \sqrt{3} } =

3

1

Tan A = \frac{1}{ \sqrt{3} } TanA=

3

1

A = 30. ------- Which is an Acute angle

Thus, the Obtuse angle between the lines = 180 - 30

= 150.

Therefore the angle between the lines is 30 or 150.

Hope this helps!

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