Question

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

Answer 1

Answer:

Step-by-step 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 $m = \frac{y_2 -y_1}{x_2-x_1}$ and $y = mx + b$ to describe the line

$m = \frac{3-(-1)}{3-(-3)} = \frac{4}{6} = \frac{2}{3} \\y = \frac{2}{3}(x) +b\\ 3 = 2*(3)/3 + b\\3 = 2 + b\\b = 1\\y = \frac{2}{3}(x)+1$

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

$m = \frac{1-(-2)}{-2-2} = \frac{3}{-4} = \frac{-3}{4} \\y = \frac{-3}{4}(x) + b\\1 = \frac{-3}{4}(-2) + b\\ 1 = 1.5 + b\\b = -\frac{1}{2} \\y = \frac{-3}{4}(x) - \frac{1}{2}$

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

$m = \frac{2+4}{-4+1} = \frac{6}{-3} = -2\\ y = -2x + b\\2 = -2(-4) + b\\2 = 8 + b\\b = -6\\y = -2x - 6$

#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)

$m = \frac{4}{1 } = 4\\ y = 4x + b\\3 = 12 + b\\b = -9\\y = 4x - 9$

#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