Concurrency of Straight Lines

 
The condition for 3 lines a1x + b1y + c1 = 0, a2x + b2y + c2 = 0, 
a3x + b3y + c3 = 0 to be concurrent is
 
        (i)     1412_Concurrency Equation 1.JPG= 0.
 
 
(ii)    There exist 3 constants l, m, n (not all zero at the same time) such that IL1 + mL2 + nL3 = 0, where L1 = 0, L2 = 0 and L3 = 0 are the three given straight lines.
 
 
(iii)    The three lines are concurrent if any one of the lines passes through the point of intersection of the other two lines.
 
Illustration:
 
        Check if lines
 
                a1 x + b1 y + c1 = 0                                             …… (1)
 
                a2 x + b2 y + c2 = 0                                             …… (2)
 
                (2a1 – 3a2)x + (2b1 – 3b2)y + (2c1 – 3c2) = 0           …… (3)
 
        are concurrent?
Solution:
        We can try to find α, ß and γ by observation as follow:
 
                L3 – 2L1 + 3L2 = 0
 
Enquiry:    many lines can pass through the intersection of two lines. Can we find a general equation of these lines?
 
If L1 = 0 and L2 = 0 are two lines then equation of family of lines passing through their intersection is given by
 
L1 + λ L2 = 0                                                       …… (A)
 
Where ‘λ’ is any parameter. (Equation A is satisfied by the point of intersection of L1= 0 and L2 = 0)
 
Note:
 
To determine a particular line one more condition is required so as to determine or eliminate λ .
 
Illustration:
 
        If x (2q + p) + y(3q + p) = 0
 
        (x + y – 1) + q/p (2x + 3y – 1) = 0, p  0
Solution:
This equation represents the family of lines passing through the intersection of lines x + y – 1 = 0 and 2x + 3y – 1 = which is fixed point i.e. (2, –1).
 
        If p = 0 then equation becomes
 
                q(2x + 3y – 1) = 0
 
        this also represents a line which passes through fixed points (2, –1).
 
Hence the given equation represents family of lines passing through a fixed point (2, –1) for variable p, q.
 
Illustration:
 
Find the equation of a line, through the intersection of 2x + 3y – 7 = 0 and x + 3y – 5 = 0 and having distance from origin as large as possible.
 

2152_concurrency of straight lines.JPG


 
Point of intersection of two lines is A(2, 1)
 
Now, with OA as radius and O itself as centre draw a circle.
 
        There will be infinitely many lines through A and each except one of them produces a chord of circle and hence their distance from origin i.e. centre of circle is less than OA i.e. radius of circle.
 
        But the exceptional one which infact is a tangent to circle at A will be at a distance OA from O.
 
        Thus, tangent to circle at A will be the line through A and is farthest from origin.
 
Now, OA  tangent at A.
 
 (slope of OA) × (slope of tangent at A) = –1
 
Or, 1–0/2–0 (slope of tangent at A) = –2
 
 equation of required line is
 
        (y – 1) = –2(x – 2)
 
Or     2x + y – 5 = 0
 
Illustration:
 
Find the point of concurrency of the altitudes drawn from the vertices (at1t2, a(t1 + r2)), (at2t3, a2t2 + t3)) and (at3t1, a(t3 + t1)) respectively of a triangle ABC.
 
 
Solution:
 
        Slope of AD = –t3.
 
        Equation of AD is y – a(t1 + t2) = –t3(x + at1t2).                … (1)
 
        Equation of CF is y – a(t3 + t1) = –t2(x – at3t1).                 … (2)
 

2146_concurrency of the altitudes.JPG

 

 
        Subtracting (1) from (2), we get
 
                x = –a Þ y = a(t1 + t2 + t1t2t3).
 
        Hence the point of concurrency of the altitudes is
 
        (–a, a(t1 + t2 + t3 + t1t2t3)).
 
 

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