# HOW TO CHECK IF PAIR OF STRAIGHT LINES INTERSECT

## About "How to Check if Pair of Straight Lines Intersect"

How to Check if Pair of Straight Lines Intersect :

Here we are going to see how to check if pair of straight lines intersect.

If the given pair of straight lines ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 intersect each other, then it must satisfy the condition given below.

abc + 2fgh - af2 - bg2 - ch2  =  0

Point of intersection :

P(hf − bg/ab − h2 , gh − af/ab − h2)

## How to Check if Pair of Straight Lines Intersect - Questions

Question 1 :

Show that the equation 2x2 −xy−3y2 −6x + 19y − 20 = 0 represents a pair of intersecting lines. Show further that the angle between them is tan−1(5).

Solution :

2x2 −xy−3y2 −6x + 19y − 20 = 0

ax2 + 2hxy + by2 + 2gx + 2fy + c = 0

a  =  2, h  =  -1/2, b  =  -3, g  =  -3, f = 19/2, c = -20

Condition for intersection of two lines :

abc + 2fgh - af2 - bg2 - ch2  =  0

2(-3)(-20)+2(19/2)(-3)(-1/2)-2(19/2)2-(-3)(-3)2-(-20)(-1/2) =  0

=  120 + 57/2 - 2(361/4) + 27 + 20(1/4)

=  120 + 57/2 - 361/2 + 27 + 5

=  (240 + 57 - 361 + 54 + 10)/2

=  (361 - 361) / 2

=  0

Since it satisfies the above condition, the given pair of straight line is intersected.

Now let us find the angle between them.

θ =  tan-1 |2√(h2- ab)/(a + b)|

θ =  tan-1 |2√(-1/2)2- 2(-3)/(2 + (-3))|

=  tan-1 |2√((1/4) + 6)/(-1)|

=  tan-1 |2√(25/4)|

=  tan-1 |(2(5)/2)|

=  tan-1 5

Hence proved.

Question 2 :

Prove that the equation to the straight lines through the origin, each of which makes an angle α with the straight line y = x is x2 − 2xy sec 2α + y2 = 0

Solution : Equation of the line AB :

y  =  m1 x

m1 x - y  =  0

Equation of the line CD :

y  =  m2 x

m12 x - y  =  0

In order to find the slope of first line, let us find the angle between x-axis and the line AB.

θ  =  45 - α

m =  tan θ  =  tan (45 - α)  ---(1)

Angle between the line CD and x-axis

θ  =  45 + α

m =  tan θ  =  tan (45 + α)  ---(2)

tan (45 - α)  =  (tan 45 - tan α)/(1 + tan 45 tan α)

m1  =  (1 - tan α)/(1 + tan α)

tan (45 + α)  =  (tan 45 + tan α)/(1 - tan 45 tan α)

m2  =  (1 + tan α)/(1 - tan α)

Equation of straight lines :

(m1 x - y) (m2 x - y)  =  0

m1m2 x2-m1xy - m2xy  + y2 =  0

m1mxxy(m1 m2) + y2 =  0   ----(3)

mm=  (1 - tan α)/(1 + tan α) + (1 + tan α)/(1 - tan α)

=  ((1 - tan α)2 + (1 + tan α)2)/(1 + tan α)(1 - tan α)

=  (1 - 2tan α + tan 2α + 1 + 2tan α + tan 2α)/(1-tan2α)

=  (2 + 2tan 2α)/(1-tan2α)

=  2[(1+tan2α)/(1-tan2α)]

=  2 sec 2α

Note :

cos 2α  =  (1-tan2α) / (1+tan2α)

mm2  =  ((1 - tan α)/(1 + tan α)) ((1 + tan α)/(1 - tan α))

=  1

By applying the above values in equation 3, we get the required equation.

(1) xxy 2 sec 2α + y2 =  0

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