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).

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