Question

find the angle between the lines x+√3y+2=0 and √3x-y+4=0​

Answer 1

Answer:

Given,

x + √3y - 2 = 0

=> √3y = -x + 2

=> y =( -1/√3 )x + 2/√3

So , here m¡ = -1/√3

And ,

√3x - y + 4 = 0

=> -y = -√3x - 4

=> y = √3x + 4

So , m¡¡ = √3

Now ,

m¡m¡¡ =√3× -1/√3

=> m¡m¡¡ = -1

So , the lines are perpendicular to each other .

Therefore , angle Between them is 90°

Happy Learning

Step-by-step explanation:

hope this answer helps you

Answer 2

Step-by-step explanation:

hey

x+√3y+2=0

√3x-y+4=0

90°