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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.
Solution
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
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