# Do the lines x+ 3y +7 =0, 3x+2y +11=0 and 2x+y+4=0 have a common point of interception?

Asked on by cruth8646

william1941 | College Teacher | (Level 3) Valedictorian

Posted on

To solve this problem, we find the point of intersection of any two lines and see if that point lies on the third line.

We have the equations:

x+ 3y +7 =0 … (1)

3x+2y +11=0 … (2)

2x+y+4=0 … (3)

(3)- 2*(1)

=> 2x+y+4 – 2*(x+ 3y +7) =0

=> 2x +y +4 – 2x-6y -14 =0

=> -5y -10 =0

=> y = -2

substituting y = -2 in (1)

x = -7 + 6 = -1

Therefore the point of contact of (1) and (3) is (-1, -2)

Putting these coordinates in (2)

3*-1 + 2*-2 +11 = -3 -4 +11 = 4, but 3x+2y +11=0.

So it does not lie on (2)

Therefore the three lines do not have a common point of intersection.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

To determine if the 3 lines have a common point, we'll have to determine the solution of the system formed by the equations of the 3 lines:

x+ 3y +7 =0

We'll subtract 7 both sides:

x + 3y = -7 (1)

3x+2y +11=0

We'll subtract 11 both sides:

3x + 2y = -11 (2)

2x+y+4=0

We'll subtract 4 both sides:

2x + y = -4 (3)

We'll determine the matrix of the system. The determinant is formed from the coefficients of x and y.

1    3

A =  3    2

2    1

We'll take the first2 lines and columns of the matrix and we'll calculate the minor of the matrix A.

1    3

d =

3     2

d = 1*2 - 3*3 = 2 - 9 = -7

Now, we'll calculate the determinant that we'll tell us if the system has solution or not.

This determinant will be formed from the minor, the last row and the column of the terms from the right side of equal:

1    3    -7

C =  3     2   -11

2    1    -4

C = -2*4 - 7*3 - 3*2*11 + 2*2*7 + 11 + 3*3*4

C = - 8 - 21 - 66 + 28 + 11 + 36

C = 20  - 1 - 30

C = -11

Since C is not zero, the system has no solutions.

If the system doesn't have solutions, that means that the lines do not intercepting or do not have any common point.

neela | High School Teacher | (Level 3) Valedictorian

Posted on

x+3y+7 = 0..........(1)

3x+2y+11 = 0...........(2)

We solve for the intersection point by solving (1) and(2):

(2)-3(1):

2y-3*3y +11-3*7 = 0

-7y-10 = 0

-7y =  10

y = -10/7.

Substituting y = -10/7 in (1) we get: x+3(-10/7) + 7 = 0

x = -7+3(10)/7 = (-49+30)/7 = -19/7.

So x = -19/7 and y = (-10/7).

We substitute  (x,y) = (-19/7 , -10/7) in the 3rd equation 2x+y+4 =0;

LHS = 2x+y+4 = 2(-19/7)+(-10/7)+4 = (-38-10)/7 +4 = -48/7+4 = (-48+28/7) = -20/7.

RHS = 0

So the intersection of the 1st 2 is not satisfying the 3rd equation.So the 3 lines are not concurrent.

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