F.4 maths

a.
L1: y = (√3)x
L2: 2(√3)x - 2y + 5 = 0

Slope of L1 = √3
Slope of L2 = -(2√3)/(-2) = √3

Since L1 and L2 have equal slopes, L1 // L2.


b.
L2: 2(√3)x - 2y + 5 = 0
When y = 0:
2(√3)x - 2(0) + 5 = 0
x = -5/(2√3) = -5(√3)/6
Hence, x-intercept of L2 = -5(√3)/6

Slope of L2 = tanθ = √3
Inclination of L2, θ = 60°


c.
L1: (√3)x - y = 0
(-5(√3)/6, 0) lies on L2.

Distance between L1 and L2
= Distance between L1 and (-5(√3)/6, 0)
= |(√3)( -5(√3)/6) - 0| / √[(√3)² + 1²]
= (5/2) / 2
= 5/4
(or 1.25)