2007 - JAMB Mathematics Past Questions and Answers - page 10

2y - kx + 2 = 0, @ x = 1, y = -2 2(-2) - k(1) + 2 = 0 -4 - k + 2 = 0 k = -2 Or y + x - \(\frac{k}{2}\) = 0, @ x= 1, y = -2 -2 + 1 - \(\frac{k}{2}\) = 0 -1 = \(\frac{k}{2}\) k = -1 x 2 = -2

L₁\(\Rightarrow\) 3Y = 4X - 1

Y = \(\frac{4}{3}\)X - \(\frac{1}{3}\)

therefore m₁ = \(\frac{4}{3}\)

L₂\(\Rightarrow\) qy = x + 3

y = \(\frac{1}{q}\)X + \(\frac{3}{q}\)

therefore, m₂ = \(\frac{1}{q}\)

For parallel lines, m₁ = m₂; \(\frac{4}{3}\) = \(\frac{1}{q}\)

therefore q = \(\frac{3}{4}\)