Question

find the value of K if the lines 3x+4y=5,2x+3y=4,Kx+4y=6are concurrent and also find the point of concurrence​

Answer 1

Answer:

Step-by-step explanation:

3x+4y=5 → (1)

2x+3y=4 → (2)

kx+4y=6 → (3)

k=3+2

k=5

Answer 2

Answer :

k = 2

Note :

• A linear equation in two variables represents a straight line on a 2D plane .
• Concurrent lines : A system of lines are said to be concurrent if all the lines of the given system intersect one another at a single point .
• If a line passes through a point then the coordinates of that point must satisfy the equation of the line .

Solution :

Here ,

The given system of linear equations is ;

3x + 4y = 5 --------(1)

2x + 3y = 4 --------(2)

kx + 4y = 6 --------(3)

Firstly ,

Let's find the point of intersection of 1st line and 2nd line .

Now ,

Multiplying eq-(1) by 2 , we get ;

=> 2•(3x + 4y) = 2•5

=> 6x + 8y = 10 -------(4)

Now ,

Multiplying eq-(2) by 3 , we get ;

=> 3•(2x + 3y) = 3•4

=> 6x + 9y = 12 --------(5)

Now ,

Subtracting eq-(4) from (5) , we get ;

=> (6x + 9y) - (6x + 8y) = 12 - 10

=> 6x + 9y - 6x - 8y = 2

=> y = 2

Now ,

Putting y = 2 in eq-(1) , we get ;

=> 3x + 4y = 5

=> 3x + 4•2 = 5

=> 3x + 8 = 5

=> 3x = 5 - 8

=> 3x = -3

=> x = -3/3

=> x = -1

Hence ,

1st line and 2nd line intersect at the point (-1 , 2) .

Also ,

It is given that , the given three lines are concurrent . Thus , they must have a common point of intersection .

Thus ,

(-1 , 2) is the point of intersection for all the given three lines .

Also ,

Since the 3rd line passes through the point (-1 , 2) , thus the coordinates of the point (-1 , 2) must satisfy the eq-(3) .

Thus ,

Putting x = -1 and y = 2 , we get ;

=> kx + 4y = 6

=> k•(-1) + 4•2 = 6

=> -k + 8 = 6

=> -k = 6 - 8

=> -k = -2

=> k = 2

Hence , k = 2 .