DETERMINE IF THE LINES ARE PARALLEL PERPENDICULAR OR NEITHER

Tell whether the lines are parallel, perpendicular, or neither.

Problem 1 :

Line 1 : through (-3, -7) and (1, 9)

Line 2 : through (-1, -4) and (0, -2)

Solution :

They are not parallel, they are not perpendicular. So, the lines are neither.

Problem 2 :

Line 1 : through (2, 7) and (-1, -2)

Line 2 : through (3, -6) and (-6, -3)

Solution :

m₁ x m₂ = 3 (-1/3)

m₁ x m₂ = -1

So, the lines are perpendicular.

Determine whether the graphs of each pair of equations are parallel, perpendicular, or neither.

Problem 3 :

y = 3x + 4

y = 3x + 7

Solution :

y = 3x + 4 ----- (1)

y = 3x + 7 ------(2)

Comparing the above equations with slope intercept form

y = mx + b, we get 

m₁ = 3, m₂ = 3

b₁ = 4, b₂ = 7

Both lines are having same slope.  but different y – intercepts. So, the given lines are parallel. 

Problem 4 :

y = -4x + 1

4y = x + 3

Solution :

y = -4x + 1 ------(1)

4y = x + 3

y = x/4 + 3/4   ------(2)

m = -4, m = 1/4

m₁ ⋅ m₂ = -4 (1/4)

m₁ ⋅ m₂ = -1

Since the product of the slopes is equal to -1. The lines are perpendicular. 

Problem 5 :

y = 2x - 5

y = 5x - 5

Solution :

y = 2x – 5   -----(1)

y = 5x - 5  -----(2)

Comparing (1) and (2)

 y = mx + b

m = 2, m = 5

b = -5, b = -5

There is no relationship between slopes, so they are neither.

Problem 6 :

y = -1/3x + 2

y = 3x - 5

Solution :

y = -1/3x + 2  -----(1)

y = 3x - 5 -----(2)

 y = mx + b

m₁ = -1/3, m₂ = 3

b = 2, b = -5

m₁ = -1/3, m₂ = 3

m₁ ⋅  m₂ = (-1/3) ⋅ 3

Product of the slope is equal to -1. So, the lines are perpendicular.

Problem 7 :

y = 3/5x - 3

5y = 3x - 10

Solution :

y = 3/5x – 3 ------(1)

5y = 3x - 10

y = 3/5x - 10 ------(2)

 y = mx + b

m₁ = 3/5, m₂ = 3/5

b = -3, b = -10

Both lines have the same slope but have different y-intercepts. So, the lines are parallel.

Problem 8 :

y = 4

4y = 6

Solution :

y = 4 ------(1)

4y = 6

y = 6/4

y = 3/2------(2)

Both are horizontal lines and they will not intersect.

Problem 8 :

The line Q passes through the points (−10, −2) and (−8, −8) The line R passes through the points (1, 2) and (10, a) The lines Q and R are perpendicular. Find the value of a.

Solution :

Slope of the line (m₁) = (y₂ - y₁)/(x₂ - x₁)

= (-8 - (-2))/(-8 - (-10))

= (-8 + 2) / (-8 + 10)

= -6/2

m₁= -3

Slope of the line (m₂) = (a - 2)/(10 - 1)

m₂= (a - 2) / 9

When lines are perpendicular, then m₁ x m₂ = -1

-3 [(a - 2) / 9] = -1

(a - 2)/3 = 1

a - 2 = 3

a = 3 + 2

a = 5

So, the value of a is 5.

Problem 8 :

The point C has coordinates (2, −3) and the point D has coordinates (4, 6). Find the equation of the line perpendicular to CD and passing through D.

Solution :

Slope of the line CD = (y₂ - y₁)/(x₂ - x₁)

= (6 - (-3)) / (4 - 2)

= (6 + 3)/2

= 9/2

The required line is perpendicular to the line CD, then slope of the required line is -2/9

The required line passes through the point D, then the equation of the line

(y - y₁) = m(x - x₁)

D(4, 6)

(y - 6) = (-2/9) (x - 4)

9(y - 6) = -2(x - 4)

9y - 54 = -2x + 8

2x + 9y = 8 + 54

2x + 9y = 62

So, the equation of the required line is 2x + 9y = 62.