Angle between Two Straight Lines:
(1) If a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 are the equations of two given lines,
the angle (theta) between the lines is given by
cos (theta) = [Modulus of ((a1*a2) + (b1*b2))] / [sqrt((a1^2+b1^)(a2^2 +b2^2))]
(2) If m1 and m2 are the slopes of two lines, the angle (theta) is
given by tan (theta) = (m1-m2)/(1+m1*m2)
(3) Condition for parallel lines is (a1/a2) = (b1/b2) or m1 = m2
(4) Condition for Perpendicular lines is (a1*a2 + b1*b2) = 0 or m1*m2 = -1
(5) The equation of a line through the point (x1, y1) & parallel to the line
ax+by+c =0 is given by a(x-x1) + b(y-y1) = 0
(6) The equation of a line through the point (x1, y1) & perpendicular to the line
ax+b+c = 0 is b(x-x1)-a(y-y1) = 0
Solved Examples:
(1) Show that the lines x+2y+3 = 0 and 2x+4y+6 = 0 are parallel
Solution:
Here a1=1, b1=2 & a2=2, b2=4
(a1/a2)= ½ & (b1/b2) = 2/4=1/2
Condition for parallelism is (a1/a2)=(b1/b2), which is equal =1/2, hence we can say that the above two lines are parallel.
(2) Show that the lines 3x+y-4 = 0 & x-3y+1 = 0 are perpendicular
Solution:
Here a1 = 3, b1 = 1 & a2 = 1, b2 = -3
The condition for perpendicular lines is (a1*a2+b1*b2) = 0
Therefore (3*1) + (1*(-3)) = 3-3 = 0, therefore we can say that the given lines are perpendicular.
(3) Find the equation of the line through the point (-1, 2) and parallel to 2x-3y+1 = 0
Solution:
Here a = 2, b = -3 & x1 = -1, y1 = 2
The required equation is a(x-x1) + b(y-y1) = 0
Substituting the values on the above we get 2(x+1) -3(y-2) = 0
On simplification we get the equation as 2x-3y+8 = 0
(4) Find the equation of the line through the point (1, 2) and perpendicular to the line
3x+2y-1 = 0
Solution:
Here a =3, b =2 & x1 =1, y1 =2
The required equation is b(x-x1)-a(y-y1) = 0
Substituting the values on the above we get 2(x-1)-3(y-2) = 0
On simplification we get the equation as 2x-3y+4 =0.
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