FINDING THE POINT OF INTERSECTION OF TWO LINES

For each of the following pair of equations find the points of intersection :

Problem 1 :

x = 5, y = -3

Solution :

x = 5, y = -3

So, the point of intersection is (5, -3).

Problem 2 :

2x + 3y = 12, x = -3

Solution :

2x + 3y = 12

When x = - 3

2(-3) + 3y = 12

3y = 12 + 6

3y = 18

y = 6

So, the point of intersection is (-3, 6).

Problem 3 :

5x – 12y = 6, y = 2

Solution :

5x – 12y = 6

When y = 2

5x - 12(2) = 6

5x - 24 = 6

5x = 6 + 24

5x = 30

x = 30/5

x = 6

So, the point of intersection is (6, -2).

Problem 4 :

y = 3x – 1, 6x – 2y = 11

Solution :

y = 3x – 1 ----(1)

6x – 2y = 11 ----(2)

6x - 2(3x - 1) = 11

6x - 6x + 2 = 11

So, no intersection point.

Problem 5 :

-3x + 5y = 11, x = 5y – 14

Solution :

-3x + 5y = 11 ----(1)

x = 5y – 14 ----(2)

Applying (2) in (1), we get

-3(5y - 14) + 5y = 11

-15y + 42 + 5y = 11

-10y = 11 - 42

-10y = -31

y = 31/10

Substitute y = 31/10 in x = 5y – 14

x = 5(31/10) – 14

= 155/10 – 14

= 15.5 – 14

x = 1.5

So, the point of intersection is (1.5, 31/10).

Problem 6 :

2x – y = 4, y = -3x - 9

Solution :

2x – y = 4 ---(1)

Applying the value of y in (1)

2x - (-3x - 9) = 4

2x + 3x + 9 = 4

5x = 4 - 9

5x = -5

x = -1

When x = -1

2(-1) - y = 4

-2 - y = 4

-2 - 4 = y

y = -6

So, the point of intersection is (-1, -6).

Problem 7 :

4x – 3y = -14, y = 2x + 5

Solution :

4x – 3y = -14 ----(1)

y = 2x + 5  ----(2)

Applying the value of y in (1), we get

4x - 3(2x + 5) = -14

4x - 6x - 15 = -14

-2x = -14 + 15

-2x = 1

x = -1/2

When x = -1/2

y = 2(-1/2) + 5

y = -1 + 5

y = 4

So, the point of intersection is (-1/2, 4).

Problem 8 :

y = x + 1, 2x – 2y = -2

Solution :

y = x + 1 ---(1)

2x – 2y = -2 ---(2)

Applying the value of y in (2), we get

2x - 2(x + 1) = -2

2x - 2x - 2 = -2

There are infinitely many solution.

Problem 9 :

y = 2x + 3, y = 5x + 6

Solution :

y = 2x + 3 --- (1)

y = 5x + 6 --- (2)

Comparing (1) and (2)

2x + 3 = 5x + 6

Comparing like terms.

2x – 5x = 6 – 3

-3x = 3

x = -1

Substitute x = -1 in y = 2x + 3

y = 2(-1) + 3

y = -2 + 3

y = 1

So, the point of intersection is (-1, 1).