Question

prove that the following lines are concurrent 5x-3y=1,2x+3y=23 and42x+21y=257​.​

Answer 1

Question:

Prove that the following lines are concurrent;

5x - 3y = 1

2x + 3y = 23

42x + 21y = 257

Note:

Concurrent lines : A set of lines are said to be concurrent if all of them intersect at a single point.

Working rule:

Step1 : Firstly, find the the point of intersection of any two of the given lines.

Step2 : Now, put the coordinates of the the obtained point in the equation of third line.

Step3 : See , whether the coordinates of the obtained point satisfies the equation of third line.

If it does, then all the given three lines can be considered as concurrent lines.

Solution:

The given set of lines are:

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

2x + 3y = 23 ---------(2)

42x + 21y = 257 -------(3)

Let's find the point of intersection of first and second line.

Thus,

Adding eq-(1) and (2) , we have;

=> 5x - 3y + 2x + 3y = 1 + 23

=> 7x = 24

=> x = 24/7

Now,

Putting x = 24/7 in eq-(1) , we get;

=> 5x - 3y = 1

=> 5•(24/7) - 3y = 1

=> 120/7 - 3y = 1

=> (120 - 21y)/7 = 1

=> 120 - 21y = 7

=> 21y = 120 - 7

=> 21y = 113

=> y = 113/21

Hence,

The point of intersection of the first and second line is (24/7,113/21).

Now,

Let's put the coordinates of the point (24/7,113/21) in the equation of third line.

ie; put x = 24/7 and y = 113/21 in eq-(3).

=> 42x + 21y = 257

=> 42•(24/7) + 21•(113/21) = 257

=> 144 + 113 = 257

=> 257 = 257

=> LHS = RHS

Thus,

It is clear that the point (24/7,113/21) satisfy the equation of third line.

Hence,

The given lines are concurrent

Answer 2

Question :-

Prove that the following lines are concurrent

5x- 3y = 1 , 2x+3y = 23

and 42x+21y = 257 .

To prove :-

→ Following lines are concurrent.

Proof :-

Procedure :-

Take any two lines and then solving both find their intersection point and then satisfy these points in that equation except we take .if they will satisfy means our assumption is correct and those lines are concurrent .

Explanation :-

Given that :

Given lines are

→ 5x- 3y = 1 ....eq.(1)

→ 2x +3y = 23 ....eq.(2)

→ 42x +21y = 257 ....eq.(3)

\________________/

So we are taking eq.(1) and eq.(3) .

Solving these to finding intersection points of these .

adding eq.(3) + 7×eq.(1)

→ 42x +21y +35x - 21y = 257+7

→ 77x = 264

→ x = 264/77

put this value of x in eq.(1)

→ 5(264/77) - 3y = 1

→ 1320/77 -1 = 3y

→ 3y = 1243/77

→ y = 1243/231

intersection points of eq.(1) and eq.(3) are ( 264/77 , 1243/231)

so satisfy these points in eq.(2)

→ 2(264/77)+3(1243/231)= 23

→ 528/77 + 1243/77 = 23

→ 1771/77 = 23

→ 23 = 23

hence satisfied.

hence prove .

\________________/