F.5 MATHS

(a)
Slope of L1 = -(1)/(-1) = 1
Slope of L2 = -(2)/(-2) = 1

Since (Slope of L1) = (Slope of L2),
thus L1 // L2.


(b)
Let (x, y) be the point.

Since L1 // L2, and the point equidistant from L1 and L2.
(x - y + 2)/√(1² + 1²) = -(2x - 2y - 5)/√(2² + 2²)
(x - y + 2)/√2 = -(2x - 2y - 5)/2√2
x - y + 2 = -x + y + 5/2
2x - 2y - (1/2) = 0
4x - 4y - 1 = 0

The locus of the point:
4x - 4y - 1 = 0

2010-11-21 16:04:31 補充：
Alternative method of (b):

Since L1 // L2,
thus (locus of the point) // L1 // L2.

y-intercept of L1 = -(2)/(-1) = 2
y-intercept of L2 = -(-5)/(-2) = -5/2

y- intercept of the locus of the point = [(2) + (-5/2)]/2 = -1/4

Locus of the point:
y = x - (1/4)
4x - 4y - 1 = 0