Question

Tell wheter the lines through the given points are parallel, perpendicular or neither.

a. Line 1: (2,-3), (-4, -6)
Line 2: (-3, 2), (2, -8)


b. Line 1: (-4, -5), (4, 1)
Line 2: ( 6, 10), (-3, -2)


c. Line 1: (0, -4), (-2, -8)
Line 2: (-1, 5), (1, 9)

Help I need this by tomorrow :c

Answer 1

Explanation:

To solve this problem, we use the following formula:

`\displaystyle\\\boxed {(x-x_1)/(x_2-x_1)=(y-y_1)/(y_2-y_1) }`

a) (2,-3) (-4,-6)

`x_1=2\ \ \ \ x_2=-4\ \ \ \ y_1=-3\ \ \ \ y_2=-6`

`\displaystyle\\(x-2)/(-4-2) =(y-(-3))/(-6-(-3)) \\\\(x-2)/(-6)=(y+3)/(-6+3) \\\\(x-2)/(-6) =(y+3)/(-3)`

Multiply both parts of the equation by -3:

`\displaystyle\\(x-2)/(2)=y+3\\\\0.5x-1=y+3\\\\0.5x-1-3=y+3-3\\\\0.5x-4=y\\`

⇔ Line 1: y=0.5x-4 m₁=0.5

(-3,2) (2,-8)

`\displaystyle\\(x-(-3))/(2-(-3))=(y-2)/(-8-2) \\\\(x+3)/(2+3) =(y-2)/(-10) \\\\(x+3)/(5) =(y-2)/(-10) \\\\`

Multiply both parts of the equation by -10:

`-2(x+3)=y-2\\-2x-6=y-2\\-2x-6+2=y-2+2\\-2x-4=y`

⇔ Line:y=-2x-4 m₂=-2

`\displaystyle\\m_1=-(1)/(m_2)`

⇒ The given lines are perpendicular

b)

(-4,-5) (4,1)

`\displaystyle\\(x-(-4))/(4-(-4))=(y-(-5))/(1-(-5)) \\\\(x+4)/(4+4)=(y+5)/(1+5) \\\\(x+4)/(8) =(y+5)/(6) \\`

Multiply both parts of the equation by 6:

`\displaystyle\\(6(x+4))/(8) =y+5\\\\(3)/(4) x+3=y+5\\\\(3)/(4) x+3-5=y+5-5\\\\(3)/(4) x-2=y\\`

⇔ Line 1: y=(3/4)x-2 m=3/4

(6,10) (-3,-2)

`\displaystyle\\(x-6)/(-3-6) =(y-10)/(-2-10) \\\\(x-6)/(-9) =(y-10)/(-12) \\\\`

Multiply both parts of the equation by -12:

`\displaystyle\\(-12(x-6))/(-9) =y-10\\\\(4)/(3)x-8=y-10 \\\\(4)/(3)x-8+10=y-10+10\\\\(4)/(3)x+2=y`

⇔ Line 2: y=(4/3)x+2 m=4/3

⇒ The given lines are neither perpendicular and neither parallel

c)

(0,-4) (-2,-8)

`\displaystyle\\(x-0)/(-2-0) =(y-(-4))/(-8-(-4)) \\(x)/(-2)=(y+4)/(-8+4) \\\\(x)/(-2)=(y+4)/(-4)`

Multiply both parts of the equation by -4:

`2x=y+4\\2x-4=y+4-4\\2x-4=y`

⇔ Line 1: y=2x-4 m₁=2

(-1,5) (1,9)

`\displaystyle\\(x-(-1))/(1-(-1)) =(y-5)/(9-5) \\\\(x+1)/(1+1) =(y-5)/(4)\\\\(x+1)/(2) } =(y-5)/(4) \\`

Multiply both parts of the equation by 4:

`2(x+1)=y-5\\2x+2=y-5\\2x+2+5=y-5+5\\2x+7=y`

⇔ Line 2: y=2x+7 m₂=2

m₁=m₂

⇒ The given lines are parallel

Answer 2

a. Perpendicular

Make sure to read the step-by-step explanation to figure out how to do the other 2.

Explanation:

I will not be providing answers to all three, but I will answer the first one (this is, so you don't depend on people to solve this for you).

Parallel lines have one thing in common, their slope. Two lines that have the same slope never intersect because they are both facing the same direction and moving the same distance. So, you can find the slope of two lines by using the slope formula.

`m=(y_2-y_1)/(x_2-x_1)\\\\m_1=(-6-(-3))/(-4-2)=(-3)/(-6)=(1)/(2)\\m_2=(-8-2)/(2-(-3))=(-10)/(5)=-2`

This is evidence that the two lines are not parallel. But, by finding the slope, we can conclude that it is perpendicular.

This brings me to how to identify perpendicular lines. The slope of perpendicular lines is negative reciprocals. A reciprocal is just a fraction that is flipped, and it's negative. For example,
`(-2)/(3)`'s negative reciprocal would be
`(3)/(2)`. The negative of -2/3 is a positive, and the reciprocal of 2/3 is 3/2. So, the negative reciprocal is 3/2.

How does this apply to what I said earlier?

Well,
`-2` is the negative reciprocal of
`(1)/(2)`. The reciprocal of
`(1)/(2)` is 2, and the negative of 2 is -2. So, simply, lines 1 and 2 are perpendicular.

Use this method to solve the other questions. If they don't have the same slope and aren't negative reciprocals, then they are neither.