Question

1. determine the distance between twogiven lines
> L1 2x - y = 5
> L2 4x - 2y = 7
2. determine the equation of the line passing through the given point ( -1,-1) and perpendicular to x -y + 3 = 0
3. determine the equation of the line passing through the given point (-2 , 5) and parallel to x + 3y - 5 = 0

FORMULAS:
absolute value of AX BY +C all over 2a (short method)
square root of (X1+X2)(Y1+Y2) (long method)

Answer 1

For L1: the y-intercept is -5
x=0 -> -y =5 -> y=-5
For L2: the y-intercept is -3.5
x=0 -> -2y = 7 -> y = -3.5

distance = |-5 -(-3.5)| = 1.5
is that what you are looking for?

Answer 2

2) x -y + 3 = 0
x+3 = y
perpendicular slope = -1
y- (-1)= -1(x-(-1) )
y+1 = -x -1
y = -x-2

Answer 3

3) x + 3y - 5 = 0
rewrite slope intercept
y = (-x +5)/3\[y=-\frac{1}{3}x+\frac{5}{3}\]
slope = -1/2
parallel than slope is the same -1/2
\[y-5= -\frac{1}{3}(x-(-2))\]\[y=-\frac{1}{2}x+\frac{13}{3}\]

Answer 4

number 1 ) it been so long I don't renember

Answer 5

tnx2.. ;D

Answer 6

np, number 1) it not hard , I don't have time go back to look the lesion

Answer 7

But you got answer number 1

Answer 8

i might be wrong, i gave the vertical distance between the lines or the vertical shift but Not the shortest distance between the 2 lines. To do that you need to find a perpendicular line that goes through both lines L1,L2 then use distance formula to find distance between the intersection points

Answer 9

number I no idea , don't renember

Answer 10

number 1) no idea