Question

Find the point of intersection of the lines 4x + 3y = 1 and 3x − y + 9 = 0. If this point lies on the line (2k – 1) x – 2y = 4; find the value of k.

Answer 1

Given lines are;

4x + 3y = 1  ...(i)

3x – y = –9  ...(ii)

Multiplying (i) by 1 and (ii) 3; we have

4x + 3y = 1

9x – 3y = –27

On Adding, we get :

13x = –26

`\implies x = -26/13 = -2`

From (ii), we have

3x – y = –9

`\implies` 3(–2) – y = –9

`\implies` −6 – y = –9

`\implies` –y = –9 + 6 = –3

∴ y = 3

∴ The point of intersection is (–2, 3)

∴ The line (2k – 1) x – 2y = 4 passes through that point (–2, 3).

∴ The point (–2, 3) satisfy it.

∴ (2k – 1)(–2) – 2(3) = 4

`\implies` −4k + 2 – 6 = 4

`\implies` −4k – 4 = 4

`\implies` −4k = 4 + 4 = 8

∴ `k = 8/(-4) = -2`

Hence point of intersection is (–2, 3) and value of k = –2