Question

Can someone please help me on 1-6, i got a bad grade in this class.

Answer 1

Explanation:

For these problems involving a straight line, all you need is two points and you can describe the line

#1 the two points are (3, 3) and (-3, -1)

we can use
`m = (y_2 -y_1)/(x_2-x_1)` and
`y = mx + b` to describe the line

`m = (3-(-1))/(3-(-3)) = (4)/(6) = (2)/(3) \\y = (2)/(3)(x) +b\\ 3 = 2*(3)/3 + b\\3 = 2 + b\\b = 1\\y = (2)/(3)(x)+1`

#2 same process. The two points given are (-2, 1) and (2, -2)

`m = (1-(-2))/(-2-2) = (3)/(-4) = (-3)/(4) \\y = (-3)/(4)(x) + b\\1 = (-3)/(4)(-2) + b\\ 1 = 1.5 + b\\b = -(1)/(2) \\y = (-3)/(4)(x) - (1)/(2)`

#3 same process, (-4, 2) and (-1, -4)

`m = (2+4)/(-4+1) = (6)/(-3) = -2\\ y = -2x + b\\2 = -2(-4) + b\\2 = 8 + b\\b = -6\\y = -2x - 6`

#4 this is a vertical line. If the m were to be calculated, it would be division by 0. These lines can be described by x = n, where n is any real number. In this case, x = 3

#5 same process as earlier (2, -1) and (3, 3)

`m = (4)/(1 ) = 4\\ y = 4x + b\\3 = 12 + b\\b = -9\\y = 4x - 9`

#6 unlike #4, this is a horizontal line, meaning the slope is zero. No matter what x value, you will always get the same number. In this case, the line is

y = -2