2007 - JAMB Mathematics Past Questions and Answers - page 10
91
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
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
Users' Answers & Comments2(-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
92
If the lines 3y = 4x - 1 and qy = x + 3 are parallel to each other, the value of q is
A
-\(\frac{4}{3}\)
B
-\(\frac{3}{4}\)
C
\(\frac{3}{4}\)
D
\(\frac{4}{3}\)
correct option: c
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 = \(\frac{3}{4}\)
Users' Answers & CommentsY = \(\frac{4}{3}\)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 = \(\frac{3}{4}\)