Question on: JAMB Mathematics - 2007
If the lines 2y - kx + 2 = 0 and y + x - \(\frac{k}{2}\) = 0 intersect at (1, 2), find the value of k
A
-2
B
-4
C
-1
D
-3
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Correct Option: A
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
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
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