Angle between pair of straight lines:
We know that the equation ax2+2hxy+by2=0 represents a pair of straight lines passing through origin and hence it can be written as product of two linear factors, ax2+2hxy+by2= (lx+my)(px+qy), where lp=a, mq=b and lq+mp=2h. Also, the separate equations of lines are lx+my=0 and px+qy=0. So, the angle between the lines is given by
Cos θ = (lp+mq)/√[(l2+m2)(p2+q2)]
= = (lp+mq)/√[(lp-mq)2 +(lq+mp)2]
= (a+b)/√[(a-b)2 + 4h2]
The above formula for Cos θ gives the acute angle between the lines ax2+2hxy+by2=0 if a+b>0 and the obtuse angle between the lines if a+b<0. Hence the acute angle between the lines represented by ax2+2hxy+by2=0 is Cos-1[|a+b|/√[(a-b)2 + 4h2]]
We know that ax2+2hxy+by2=0=0 represents a pair of parallel lines (coincident) if h2=ab. Now, the lines represented by ax2+2hxy+by2=0 are perpendicular if Cos θ=0. i.e. the lines are perpendicular of a+b=0.
(i.e. the lines are perpendicular if the sum of coefficients of x2 and y2 is zero).
If a+b ≠0, then the lines represented by ax2+2hxy+by2=0 are not perpendicular nad the angle between them is represented by either of these formulas.
Tan θ = [2√(h2-ab)]/(a+b)
Cos θ = (a+b)/√[(a-b)2 + 4h2]
Sin θ = [2√(h2-ab)]/√[(a-b)2 + 4h2]
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