consider line 4x-3y+12=0

Question

consider line 4x-3y+12=0,find the equation of the line passing through the point 1,2 and the parallel to the given line. and find the distance between these two parallel lines

Answer 1

4x-3y+12=0

Rearranging to standard format of ax + by = c

4x - 3y = -12

 

The slope of ax + by = c is given by m = -a/b

m = -4/-3 = 4/3

Parallel lines have same slope.  So both lines must have slope m = 4/3.

 

Equation of line with slope 4/3 through (1, 2)

 

y = mx + b

y = (4/3)x + b

Using (1,2) as (x, y) find b

2 = (4/3)(1) + b

b = 2 - 4/3 = 2/3

y = (4/3)x + 2/3

 

We can clear the fractions by multiplying through by 3

3y = 4x + 2

-4x + 3y = 2

 

We have two parallel lines:

4x - 3y = -12   L1

-4x + 3y = 2    L2

 

We want the perpendicular distance between the two lines.

Find the equation of the line through (1,2) that is perpendicular to

4x - 3y = -12.

As already established, m = 4/3 for this line.  Perpendicular line

will have slope -1/m = -3/4

Line through (1, 2) with slope -3/4

y = (-3/4)x + b

2 = (-3/4)(1) + b

b = 2 + 3/4 = 11/4

y = (-3/4)x + 11/4

4y = -3x + 11

3x + 4y = 11    L3

 

We now have the equation of a line perpendicular to L2 which

meets it at point (1,2)

Now find the intersection point of this perpendicular line with

the 1st equation L1

 

4x - 3y = -12    L1

3x + 4y = 11    L3

 

multiply L1 by 4 and L2 by 3 and add equations

16x - 12y = -48

9x + 12y = 33

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

25x = -15

x = -15/25 = -3/5

 

4x - 3y = -12   

4(-3/5) - 3y = -12

-12/5 - 3y = -12

-3y = -12 + 12/5

-3y = -48/5

y = -48/-15 = 3 1/5 = 16/5

 

The intersection point is (-3/5, 16/5)

 

The distance from (1, 2) to (-3/5, 16/5) is the perpendicular distance

between the two parallel lines

 

d = √[(x₂-x₁)² + (y₂-y₁)²]

d = √[(1-(-3/5))²+(2-16/5)²]

d = √[(8/5)² + (-6/5)²]

d = √(64/25 + 36/25)

d = √(100/25) = 10/5 = 2