Concurrency of Straight LinesThe condition for 3 lines a_{1}x + b_{1}y + c_{1}= 0, a_{2}x + b_{2}y + c_{2}= 0,

a_{3}x + b_{3}y + c_{3}= 0 to be concurrent is(i) = 0.(ii) There exist 3 constants l, m, n (not all zero at the same time) such that IL_{1}+ mL_{2}+ nL_{3}= 0, where L_{1}= 0, L_{2}= 0 and L_{3}= 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 linesa_{1}x + b_{1}y + c_{1}= 0 …… (1)a_{2}x + b_{2}y + c_{2}= 0 …… (2)(2a_{1}– 3a_{2})x + (2b_{1}– 3b_{2})y + (2c_{1}– 3c_{2}) = 0 …… (3)are concurrent?Solution:We can try to find α, ß and γ by observation as follow:L_{3}– 2L_{1}+ 3L_{2}= 0Enquiry: many lines can pass through the intersection of two lines. Can we find a general equation of these lines?If L_{1}= 0 and L_{2}= 0 are two lines then equation of family of lines passing through their intersection is given byL_{1}+ λ L_{2}= 0 …… (A)Where ‘λ’ is any parameter. (Equation A is satisfied by the point of intersection of L_{1}= 0 and L_{2}= 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 ≠ 0Solution: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 becomesq(2x + 3y – 1) = 0this 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.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) = –1Or, 1–0/2–0 (slope of tangent at A) = –2∴ equation of required line is(y – 1) = –2(x – 2)Or 2x + y – 5 = 0Illustration:Find the point of concurrency of the altitudes drawn from the vertices (at_{1}t_{2}, a(t_{1}+ r_{2})), (at_{2}t_{3}, a_{2}t_{2}+ t_{3})) and (at_{3}t_{1}, a(t_{3}+ t_{1})) respectively of a triangle ABC.Solution:Slope of AD = –t_{3}.Equation of AD is y – a(t_{1}+ t_{2}) = –t_{3}(x + at_{1}t_{2}). … (1)Equation of CF is y – a(t_{3}+ t_{1}) = –t_{2}(x – at_{3}t_{1}). … (2)Subtracting (1) from (2), we getx = –a Þ y = a(t_{1}+ t_{2}+ t_{1}t_{2}t_{3}).Hence the point of concurrency of the altitudes is(–a, a(t_{1}+ t_{2}+ t_{3}+ t_{1}t_{2}t_{3})).AskIITians offers a novel way of teaching where you can prepare for IIT JEE, AIEEE and other engineering examinations for free by sitting at home. You can visit the website askIITians.com to read the study material pertaining to your preparation. Co-ordinate Geometry is handled in detail in the study material at askIITians.com. Be a part of our online tests and AQAD (A Question A Day) for free and be a winner.