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Concurrency of Straight Lines What do you mean by intersection of three lines or concurrency of straight lines? Suppose we have three staright lines whose equations are a_{1}x + b_{1}y + c_{1} = 0, a_{2}x + b_{2}y + c_{2} = 0 and a_{3}x + b_{3}y + c_{3} = 0. These lines are sid to be concurrent if the following condition holds good: = 0. If we have three straight lines with equations L_{1} = 0, L_{2} = 0 and L_{3} = 0, then they are said to be concurrent if there exist three constants a, b and c not all zero such that aL_{1} + bL_{2} + cL_{3} = 0. Generally speaking, three lines are said to be concurrent if any one of the lines passes through the point of intersection of the other two lines. Now, we discuss certain illustrations based on this concept: Illustration: Check if the following lines are concurrent a_{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) Solution: On carefully seeing the coeffcients we realize that if the lines are denoted by L_{1}, L_{2} and L_{3}, then we have L_{3} – 2L_{1} + 3L_{2} = 0. Hence, using the above fact, we have found three constants not all zero such that aL_{1} + bL_{2} + cL_{3} = 0. Hence, the lines are concurrent. Result: How 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 by L_{1} + λL_{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 . You may also view the video on intersection of straight lines Illustration: Find the equation of a line passing through the point of intersection of x + y – 3 = 0 and 2x – y + 1 = 0 and a point (2, -3). Solution: Any line passing through the point of intersection of the lines x + y – 3 = 0 and 2x – y + 1 = 0 is given by (x + y – 3) +A(2x – y + 1) = 0 Now, it passes through the point (2, -3) and so (2-3-3) + A(2.2 – (-3) + 1) = 0 This gives 8A – 4 = 0 which gives A = 1/2 Hence, the required equation is (x + y – 3) + 1/2(2x – y + 1) = 0 which is 4x + y – 5 = 0. 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. Solution: 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. Hence, (slope of OA) × (slope of tangent at A) = –1 Or, (slope of tangent at A) = –2 which means (1–0)/(2–0) = -2. Hence, 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 (at_{1}t_{2}, a(t_{1} + t_{2})), (at_{2}t_{3}, a(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 – at3t_{1}). … (2) Subtracting (1) from (2), we get x = –a which gives 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})). Related Resources:
Suppose we have three staright lines whose equations are a_{1}x + b_{1}y + c_{1} = 0, a_{2}x + b_{2}y + c_{2} = 0 and a_{3}x + b_{3}y + c_{3} = 0. These lines are sid to be concurrent if the following condition holds good:
= 0.
If we have three straight lines with equations L_{1} = 0, L_{2} = 0 and L_{3} = 0, then they are said to be concurrent if there exist three constants a, b and c not all zero such that aL_{1} + bL_{2} + cL_{3} = 0.
Generally speaking, three lines are said to be concurrent if any one of the lines passes through the point of intersection of the other two lines.
Now, we discuss certain illustrations based on this concept:
Check if the following lines are concurrent
a_{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)
On carefully seeing the coeffcients we realize that if the lines are denoted by L_{1}, L_{2} and L_{3}, then we have L_{3} – 2L_{1} + 3L_{2} = 0. Hence, using the above fact, we have found three constants not all zero such that aL_{1} + bL_{2} + cL_{3} = 0. Hence, the lines are concurrent.
How 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 by
L_{1} + λL_{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)
To determine a particular line, one more condition is required so as to determine or eliminate λ .
You may also view the video on intersection of straight lines
Find the equation of a line passing through the point of intersection of x + y – 3 = 0 and 2x – y + 1 = 0 and a point (2, -3).
Any line passing through the point of intersection of the lines x + y – 3 = 0 and 2x – y + 1 = 0 is given by
(x + y – 3) +A(2x – y + 1) = 0
Now, it passes through the point (2, -3) and so (2-3-3) + A(2.2 – (-3) + 1) = 0
This gives 8A – 4 = 0 which gives A = 1/2
Hence, the required equation is (x + y – 3) + 1/2(2x – y + 1) = 0 which is 4x + y – 5 = 0.
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.
Hence, (slope of OA) × (slope of tangent at A) = –1
Or, (slope of tangent at A) = –2 which means (1–0)/(2–0) = -2.
Hence, equation of required line is (y – 1) = –2(x – 2)
or 2x + y – 5 = 0.
Find the point of concurrency of the altitudes drawn from the vertices (at_{1}t_{2}, a(t_{1} + t_{2})), (at_{2}t_{3}, a(t_{2 }+ t_{3})) and (at_{3}t_{1}, a(t_{3} + t_{1})) respectively of a triangle ABC.
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 – at3t_{1}). … (2)
Subtracting (1) from (2), we get
x = –a which gives 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})).
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