Finding the Point of Intersection of Two Lines Examples :
If two straight lines are not parallel then they will meet at a point.This common point for both straight lines is called the point of intersection.
If the equations of two intersecting straight lines are
given then their intersecting point is obtained by solving equations
simultaneously.
Example 1 :
Find the intersection point of the straight lines
3 x + 5 y - 6 = 0 and 5x - y - 10 = 0
Solution :
3x + 5y - 6 = 0 ----- (1)
5x - y - 10 = 0 ------(2)
(2) ⋅ 5 ==> 25 x - 5 y - 50 = 0
3x + 5y - 6 = 0
25x - 5y - 50 = 0
------------------
28 x - 56 = 0
28x = 56
x = 2
By applying x = 2 in (1), we get
3x + 5y - 6 = 0
3(2) - 6 + 5 y = 0
6 - 6 + 5 y = 0
5y = 0
y = 0
So, the point of intersection of the straight lines is (2, 0).
Example 2 :
Find the intersection point of the straight lines
2x + 3y = 5 and 3x + 4y = 7
Solution :
2x + 3y = 5 ----- (1)
3x + 4y = 7 ------(2)
(1) ⋅ 4 => 8x + 12y = 20
(2) ⋅ 3 => 9x + 12y = 21
8 x + 12 y = 20
9 x + 12 y = 21
(-) (-) (-)
--------------------
- 1 x = - 1
x = 1
By applying x = 1 in (1), we get
8(1) + 12y = 20
8 + 12 y = 20
12 y = 20 - 8
12 y = 12
y = 1
So the point of intersection of the straight lines is (1, 1).
Example 3 :
Find the intersection point of the straight lines
5x + 6y = 25 and 3x - 4y = 10
Solution :
5x + 6y = 25 ----- (1)
3x - 4y = 10 ------(2)
(1) ⋅ 2 => 10 x + 12 y = 50
(2) ⋅ 3 => 9 x - 12 y = 30
10x + 12y = 50
9x - 12y = 30
--------------------
19 x = 80
x = 80/19
By applying x = 80/19 in (1), we get
5 (80/19) + 6 y = 25
400/19 + 6 y = 25
(400 + 114y)/19 = 25
400 + 114 y = 25 ⋅ 19
400 + 114 y = 475
114 y = 475 - 400
114 y = 75
y = 75/114
y = 25/38
So, the point of intersection of the straight lines is
(80/19, 25/38).
After having gone through the stuff given above, we hope that the students would have understood how to find the point of intersection of two lines.
Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
Apr 20, 24 12:02 AM
Apr 19, 24 11:58 PM
Apr 19, 24 11:45 PM