# If the lines: 2x+y-3=0, 5x+ky-3=0 and 3x-y-2=0 are concurrent, find the value of 'k'?

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asked Aug 18, 2017

## 1 Answer

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answered Dec 8, 2018 by Pro (116 points)
We know that, Three lines are said to be concurrent, if they pass through a common point which also means that the point of intersection of any two lines always lies on the third line.

Now, the three given equations of line are:

2x + y − 3 = 0 ... (1)

5x + ky − 3 = 0 ... (2)

and 3x − y − 2 = 0 ... (3)

Firstly, we will solve the equations (1) and (3) and find out the value of x and y. The corresponding value of x and y will also satisfy the equation (2) as all the given lines are concurrent.

Therefore, on adding (1) and (3), we get

2x + y − 3 + 3x − y − 2 = 0

⇒ 5x - 5 = 0

⇒ x = 1

On putting value of x in (1), we get

2(1) + y − 3 =0

⇒ y - 1 = 0

⇒ y = 1

So, the point of intersection of two lines is (1, 1).

Now, this point will also satisfy the equation (2).

Therefore, on putting (x, y) = (1, 1) in (2), we get

5(1) + k(1) − 3 = 0

⇒ k + 2 = 0

⇒ k = - 2

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