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Step-by-step explanation:

Number 1

For parallelism

gradient of the two lines is the same

y = 3/2x -4

y = mx + C

m = 3/2

(-1, 0) = (x1, y1)

y - y1 = m(x - x1)

y - 0 = 3/2[x - (-1)]

y = 3/2(x + 1)

2y = 3x + 3

y = 3/2x + 3/2

Option A

Number 2

For parallelism

gradient of the two lines is the same

y = -½x + 2

y = mx + c

m = -½

(-2, -1) = (x1, y1)

y - y1 = m(x - x1)

y - (-1) = -½[x - (-2)]

y + 1 = -½(x + 2)

2(y + 1) = -(x + 2)

2y + 2 = -x - 2

2y = -x - 2 - 2

2y = -x - 4

y = -½x - 2

Option A

Number 3

For perpendicularity... m1 × m2 = -1

y = 4/3x - 5

y = mx + c

m1 = 4/3

m2 = -1/m1

m2 = -1/(4/3)

m2 = -3/4

(3, -1) = (x1, y1)

y - y1 = m2(x - x1)

y - (-1) = -¾(x - 3)

y + 1 = -¾(x - 3)

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

4y + 4 = -3x + 9

4y = -3x + 9 - 4

4y = -3x + 5

y = -¾x + 5/4

Option A

Number 4

For perpendicularity... m1 × m2 = -1

y = -5x + 5

y = mx + c

m1 = -5

m2 = -1/m1

m2 = -1/-5

m2 = ⅕

(5, 4) = (x1, y1)

y - y1 = m(x - x1)

y - 4 = ⅕(x - 5)

5(y - 4) = x - 5

5y - 20 = x - 5

5y = x - 5 + 20

5y = x + 15

y = ⅕x + 15/5

y = ⅕x + 3

Option C

-14 is the answer

hope this : ))

yes

step-by-step explanation:

yes