Solved Examples on Concurrent Lines
Example 1: Prove the following set of three lines are concurrent. 15x – 18y + 1 = 0, 12x + 10y -3 = 0 and 6x + 66y – 11 = 0.
Solution:
Given:
15x – 18y + 1 = 0 …. (i)
12x + 10y – 3 = 0 ….(ii)
6x + 66y – 11 = 0 ….(iii)
Using Method 1
Determinant form is given as [Tex]\begin{vmatrix} 15 & -18 & 1\\ 12 &10 &-3 \\ 6& 66 & -11 \end{vmatrix} [/Tex]
Solving the above determinant
[Tex]15\begin{vmatrix} 10& -3\\ 66 & -11 \end{vmatrix} -(-18)\begin{vmatrix} 12& -3\\ 6 & -11 \end{vmatrix} +1\begin{vmatrix} 12& 10\\ 6 & 66 \end{vmatrix} [/Tex]
= 15(-110 + 198) + 18(-132 + 18) + 1(792-60)
= 0 .
Hence Proven, lines are concurrent.
Example 2: Find the value of c for which the three lines are concurrent 2x – 5y + 3 = 0, 5x – 9y + c = 0, x – 2y + 1 = 0 .
Solution:
Given
2x – 5y + 3 = 0 ……(i)
5x – 9y + c = 0 ……(ii)
x – 2y + 1 = 0 ……(iii)
(Using Method 2)
Step 1: solve equations (i) and (iii) using the substitution method
2x – 5y + 3 = 0
x – 2y + 1 = 0
upon solving we get x = 1 and y = 1
Step 2:Substitute the values in equation (ii).
5(1) – 9(1) + c = 0
c = 4
Example 3: Prove the following set of three lines are concurrent M1 = (a-b)x + (b-c)y + (c-a) = 0, M2 = (b – c)x + (c-a)y + (a-b) = 0, M3 = (c-a)x + (a-b)y + (b-c) = 0.
Solution:
Given:
M1 = (a – b)x + (b – c)y + (c – a) = 0 ….(i)
M2 = (b – c)x + (c – a)y + (a – b) = 0 ….(ii)
M3 = (c – a)x + (a – b)y + (b – c) = 0. ….(iii)
Determinant form,
[Tex]\begin{vmatrix} (a-b) &(b-c) & (c-a)\\ (b-c)& (c-a) & (a-b)\\ (c-a)&(a-b) & (b-c) \end{vmatrix} [/Tex]
= (a – b)[(c – a)(b – c) – (a – b)(a – b)] – (b – c)[(b – c)(b – c) – (c – a)(a – b)] + (c – a)[(b -c)(a – b) – (c-a)(c – a)]
= (a – b)(c – a)(b – c) – (a – b)3 – (b – c)3 + (b – c)(c – a)(a – b) + (c – a)(b – c)(a – b) – (c – a)3
= 3 × (a – b)(b – c)(c – a) – (a – b)3 – (b-c)3 – (c-a)3
= 3 × (a – b)(b – c)(c – a) – 3a2(c − b) – 3b2(a − c) – 3c2(b − a)
= 0
Hence Proven, lines are concurrent.
Example 4: The following set of three lines are concurrent a1x + b1y + 1 = 0, a2x + b2y + 1 = 0, and a3x + b3y + 1 = 0. Prove that the points (a1 , b1), (a2 , b2), (a3 , b3) are collinear.
Solution:
If the following sets of lines are concurrent then there determinant must be
[Tex]\begin{vmatrix} a_{1} & b _{1}& 1\\ a_{2}& b_{2} & 1\\ a_{3}& b_{3} & 1 \end{vmatrix} = 0 [/Tex]
Now the above is also the condition for collinearity. Hence the given points (a1 , b1), (a2 , b2) , (a3 , b3) are collinear.
Concurrent Lines
Concurrent Lines occur when three or more lines intersect at a single point. This concept is a significant topic within the realm of straight lines.
In this article on Concurrent Lines, we will delve into the precise definition of concurrent lines, explore the conditions that lead to concurrent lines, examine the role of concurrent lines within triangles, distinguish concurrent lines from intersecting lines, and solve some practice problems to get a better understanding of the topic.
Table of Content
- Meaning of Concurrent Lines
- Examples of Concurrent Lines
- Condition for three lines to be Concurrent
- Concurrent Lines, Line Segments and Rays
- Concurrent Lines Vs Intersecting Lines
- Concurrent Lines in a Triangle