Definition of intersecting lines:
Consider the equations
a1x+b1y+c1 = 0
a2x+b2y+c2 =0
These equations represent two straight lines. If a point P lies
either in any equation then P satisfies the equation
(a1x+b1y+c1 = 0) (a2x+b2y+c2)
=0→ (3)
Conversely if P is a point satisfying the equation (3), then P
satisfies either (1) or (2) and hence p lies either (1) or (2). Thus equation
(3) represents both the lines (1) and (2).By expanding the equation (3) it can
be reduced to the form ax2+2hxy+by2+2gx+2fy+c=0 a second
degree linear equation in x and y.
Consequently the combined equation of the two lines a a1x+b1y=0,
a2x+b2y=0 which passes through the origin is (a1x+b1y)(a2x+b2y)
=0 and it is of the form ax2+2hxy+by2 = 0, a second
degree homogeneous equation in x and y.
1)If h2≥ ab,
then ax2+2hxy+by2 = 0 represents a pair of straight
lines passing through the origin.
Note:- If h2<
ab then ax2+2hxy+by2 = 0 represents two imaginary
lines having real point of intersection, the origin.
Note 2:- If h2=
ab then ax2+2hxy+by2 = 0 represents coincident
lines.
Note 3:- h2>
ab then ax2+2hxy+by2 = 0 represents two real and
different lines.
Intersecting lines:- If
the two pair of lines are intersects at a point is called the pair of
intersecting lines.
Angle between a pair of
lines:- If ө is an angle between
the lines represented by ax2+2hxy+by2 = 0 then cos ө
= (a+b)/√((a-b)2+4h2
1)If ө be the acute
angle between the lines represented by ax2+2hxy+by2 =
0, then cos ө =( mod of a+b)/ √((a-b)2+4h2
2) If ө be the acute
angle between the lines represented by ax2+2hxy+by2 =
0, then Tan ө = 2√(h2-ab)
3) If ө=0, then the two
lines will be coincident and tanө = 0 which implies h2-ab =0
which implies h2=ab
3)If ө= ∏ /2, then the
two lines will be perpendicular and cos ө = 0 which implies a+b =0 which
implies co efficient of x2+coefficient of y2
More on Definition of Intersecting Lines
Angle bisectors:- The equations of bisectors of angle
between the lines a1x+b1y+c1 = 0, a2x+b2y+c2 =0
are
(a1x+b1y+c1)/
√(a12+b12) (+ or -)
(a2x+b2y+c2 )/ √(a22+b22)
=0 If c1c2>0 then the line bisection the angle
containing the origin between the lines is
(a1x+b1y+c1)/
√(a12+b12) -
(a2x+b2y+c2 )/ √(a22+b22)
=0 and the line bisecting the other angle is
(a1x+b1y+c1)/
√(a12+b12) + (a2x+b2y+c2 )/
√(a22+b22) =0
Pair of angle
bisectors:- The equation to
the pair of bisectors between the pair of lines ax2+2hxy+by2 =
0 is h(x2-y2) = (a-b)xy
DEFINITION:-A pair of lines L1L2 =
0 is said to be equally inclined to a line L =0 if the lines L1 =0,
and L2=0 subtend the same angle with the line L =0
1)The equation to the
pair of lines passing through the origin and forming an equilateral triangle
with the line ax+by+c=0 is (ax+by)2- 3(bx-ay)2=0
Solved Problems on Definition of Intersecting Lines
Q 1: Find the equations of bisectors of the
angle between the lines x+y-2 =0 and x-7y+5 =0 and distinguish them.
Sol :- Equations of the bisectors of the angle between
the given lines are (x+y-2) / √(1+1) (+ or -) (x-7y+5)/ √(1+49) =0
(x+y-2) / √2 (+ or -) (x-7y+5)/ 5√2 = 0
5(x+y-2)(+ or -) 1(x-7y+5) = 0
5x+5y-10+x-7y+5 = 0 or
5x+5y-10-x+7y-5 = 0
6x-2y-5 =0 or 4x+12y-15
= 0
If ө is the acute angle
between x+y-2 =0, 6x-2y-5 =0 then
Cos ө = mod of (6-2)/( √(1+1) √(36+4) = 4/4√5 = 1/√5<1/√2
= cos (∏ /4)
= ө >∏ /4
Therefore 6x-2y-5 =0 is
the obtuse angle bisector and hence 4x+12y-15 =0 is the acute angle bisector.
Q 2: Find the
equation to the pair of lines passing through the origin and forming an
equilateral triangle with the line 2x+3y+5 =0 Also find the area.
Sol :- The equation to the required parallel
lines is
(2x+3y)2-3(3x-2y)2 =0
(4x2+9y2+12xy)-3(9x2+4y2-12xy)
= 0
-23x2+48xy-3y2 =
0
23x2-48xy+3y2 =
0
Therefore the length of
altitude p = 5/ √(4+9) = 5/√(13)
Area of the triangle = P2/√3
= 25/13√3 sq. unit
