equations of st. line (5m)

Question 1:

First you find the slope of equation L as you need the slope of L for equation L1. Since two lines are parallel, the two slopes should be the same.
You don’t need to find the general form of equation L, just the slope is good enough.
Consider the x-intercept (2, 0) and y-intercept (0, 3) of equation L
m = (y2 – y1)/(x2 – x1) = ( 3-0)/(0-2) = - 3/2

For equation L1, you know the x-intercept (1/2, 0) and the slope = -3/2
The equation is given by (y – y1) = m (x – x1) where (x1, y1) is any point on the straight line. We use the point (1/2, 0)
y - 0 = -3/2 (x – ½)
y = = -3/2 x + ¾
4y = -6x + 3

4y + 6x – 3 = 0 (general form)

Question 2:

Slope of line L1 = -(a -1) /(2a + 3)
Slope of line L2 = -a/(1-a)
Since these two straight lines are perpendicular to each other
Slope of L1 X Slope of L2 = -1
-(a -1) - (a)
-------- X -------------- = -1
2a + 3 (1 – a)

(1 - a) - (a)
-------- X -------------- = -1
2a + 3 (1 – a)

-(1 – a) (a) = - (1 - a) (2a + 3) ----------------------- (1)
- a +a^2 = 2a^2 + 2a - 3
2a^2 + a - 3 + a –a^2 = 0
a^2 + 2a - 3 = 0
(a – 1)(a + 3) = 0

a = 1 or a = -3

You get one answer because you cancel out the factor (1- a) on both sides of equation (1)