Question

Find the angles between the lines
x+√3y-2 =0 and √3x-y +4 =0​

Answer 1

Here's the Solution :

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°

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