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

${\large\bf{\mid{\overline{\underline{To\:Prove:-}}}\mid}}$

• 5x-3y = 1
• 2x+3y = 23
• 42x+21y = 257
• Prove that these lines are concurrent ?

$\Large\bold\star\underline{\underline\textbf{Concept\:used}}$

• Solve any two equations of the straight lines and obtain their point of intersection.
• Plug the co-ordinates of the point of intersection in the third equation.
• If it is satisfied, the point lies on the third line and so the three straight lines are concurrent.

____________________________

$\Large\underline{\underline{\sf{Solution}:}}$

Let,,

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

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

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

$\textbf{we will Try to solve Eq.(1) and Eq.(2) First}$

_______________________

Adding Equation (1) and Equation (2) we get,,

→ (5x - 3y) + (2x + 3y) = 23 + 1

→ 7x = 24

$\red{\boxed\implies} \: \large\boxed{\bold{x = \frac{24}{7} }}$

Putting in Equation (1) we get,

$\red\leadsto \: 5 \times \frac{24}{7} - 3y = 1 \\ \\ \red\leadsto \: \frac{120}{7} - 1 = 3y \\ \\ \red\leadsto \: \frac{120 - 7}{7} = 3y \\ \\ \red\leadsto \: \frac{113}{7} = 3y \\ \\ \red\leadsto \: y = \frac{113}{7 \times 3} \\ \\ \red\leadsto \: \large\boxed{\bold{y= \frac{113}{21} }}$

we get, both values of x and y ....

__________________________

Now, lets check if they satisfy in Equation (3) or not ,,,

$42x + 21y = 257 \\ \\ \red\leadsto \: \cancel{42} \times \frac{24}{\cancel{7}} + \cancel{21} \times \frac{113}{ \cancel{21}} = 257 \\ \\ \red\leadsto \: 6 \times 24 + 113 = 257 \\ \\ \red\leadsto \: 144 + 113 = 257 \\ \\ \red\leadsto \: \red{\large\boxed{\bold{257 = 257}}}$

______________________________

Clearly, the point of intersection of the lines (i) and (ii) satisfies the third equation.

$\textbf{Hence, Lines Are concurrent .}$

$\large\underline\textbf{Hope it Helps You.}$