In each, prove that the three lines are concurrent and also find the point of concurrency.
Problem 1 :
2x + y = 1
2x + 3y = 3
3x + 2y = 2
Problem 2 :
x + y - 5 = 0
3x - y + 1 = 0
5x - y - 1 = 0
Problem 3 :
3x + y = -2
2x - y = -3
x + 4y = 3
1. Answer :
2x + y = 1 ----(1)
2x + 3y = 3 ----(2)
3x + 2y = 2 ----(3)
Solve (1) and (2) to find the point of intersection.
Solve (1) for y in terms of x.
2x + y = 1
y = 1 - 2x ----(4)
Substitute y = 1 - 2x into (2).
2x + 3(1 - 2x) = 3
2x + 3 - 6x = 3
-4x + 3 = 3
-4x = 0
x = 0
Substitute x = 0 into (4).
y = 1 - 2(0)
y = 1 - 0
y = 1
The point of intersection (1) and (2) is (0, 1).
Substitute (0, 1) into (3).
3(0) + 2(1) = 2 (true ?)
0 + 2 = 2 (true ?)
2 = 2 (true)
The point (0, 1) satisfies (3).
Therefore, the given three lines are concurrent and the point of concurrency is (0, 1).
2. Answer :
x + y - 5 = 0 ----(1)
3x - y + 1 = 0 ----(2)
5x - y - 1 = 0 ----(3)
Solve (1) and (2) to find the point of intersection.
Solve (1) for y in terms of x.
x + y - 5 = 0
y = 5 - x ----(4)
Substitute y = 5 - x into (2).
3x - (5 - x) + 1 = 0
3x - 5 + x + 1 = 0
4x - 4 = 0
4x = 4
x = 1
Substitute x = 1 into (4).
y = 5 - 1
y = 4
The point of intersection (1) and (2) is (1, 4).
Substitute (1, 4) into (3).
5(1) - 4 - 1 = 0 (true ?)
5 - 4 - 1 = 0 (true ?)
0 = 0 (true)
The point (1, 4) satisfies (3).
Therefore, the given three lines are concurrent and the point of concurrency is (1, 4).
3. Answer :
3x + y = -2 ----(1)
2x - y = -3 ----(2)
x + 4y = 3 ----(3)
Solve (1) and (2) to find the point of intersection.
Solve (1) for y in terms of x.
3x + y = -2
y = -2 - 3x ----(4)
Substitute y = -2 - 3x into (2).
2x - (-2 - 3x) = -3
2x + 2 + 3x = -3
5x + 2 = -3
5x = -5
x = -1
Substitute x = -1 into (4).
y = -2 - 3(-1)
y = -2 + 3
y = 1
The point of intersection (1) and (2) is (-1, 1).
Substitute (-1, 1) into (3).
x + 4y = 3 (true ?)
-1 + 4(1) = 3 (true ?)
-1 + 4 = 3 (true ?)
3 = 3 (true)
The point (-1, 1) satisfies (3).
Therefore, the given three lines are concurrent and the point of concurrency is (-1, 1).
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
Dec 01, 23 08:59 PM
Dec 01, 23 11:01 AM
Nov 30, 23 04:36 AM