Line ℓ1 has the equation - x+y= - 3 and line ℓ2 has the equation y=x+1. Find the distance between ℓ1 and ℓ2.

Question

finding the distance between two parallel lines, round answer to the nearest tenth

Answer 1

L1:  - x+y= - 3 ...  y = x-3

L2:  y = x + 1

 

The distance will be the length of a line segment along

a perpendicular line that intersects L1 and L2

 

y-intercept of L1 is at (0, -3)

slope of L1 is 1

Perpendicular slope is -1/1 = -1

 

We will find the equation of the line through

(0, -3) perpendicular to y = x-3 

Once we have that we need to find the intersection

point with that line and L2 which is y = x + 1.

The distance we need is the distance from (0, -3)

to this intersection point.

 

We need a line through (0, -3) with slope -1

y = mx + b

y = -x + b

-3 = -(0) + b

b = -3

y = -x - 3  

This line is perpendicular to the original lines L1 and L2

and goes through (0, -3) on L1

 

We now need to find the intersection point with that

line and L2

 

y = -x - 3

y = x + 1    add the equations

---------------

2y = -2

y = -1

Substitute that into one of the equations to find x:

-1 = x + 2

x = -3

 

The intersection point is (-3, -1)

 

Now just use the distance formula to find the

distance between (0, -3) and (-3, -1)

 

d = √[(x₂-x₁)²+(y₂-y₁)²]

 

d = √[(-3-0)²+(-1-(-3))²]

 

d = √(9 +4)

 

d = √13

approximately 3.6

Answer 2

4

 

The y intercepts are -3 and 1