Question

2.1.2 Linear Tables and Graphs (Algebra 1)
The question is on the image below

Answer 1

Explanation:

1). Let the equation of a line is,

y = mx + b

Here, m = slope of the line

b = y-intercept

From the graph attached,

Slope =
`\frac{\text{Rise}}{\text{Run}}` =
`(1)/(4)`

y-intercept = -2

Equation → y =
`(1)/(4)x-2`

2). Slope of the line =
`\frac{\text{Rise}}{\text{Run}}=(3)/(4)`

y-intercept 'b' = -3

Equation → y =
`(3)/(4)x-4`

3). Slope of the line =
`\frac{\text{Rise}}{\text{Run}}=(-5)/(0.75)`

=
`-(20)/(3)`

y-intercept = 5

Equation → y =
`-(20)/(3)+5`

4). Slope of the line =
`\frac{\text{Rise}}{\text{Run}}=(-2)/(4)=-(1)/(2)`

y-intercept = 4

Equation → y =
`-(1)/(2)x+4`

5). Since, line is passing through origin (0, 0)

y-intercept = 0

Slope =
`\frac{\text{Rise}}{\text{Run}}=(1)/(1)`

Equation → y = x

6). Slope of the line =
`\frac{\text{Rise}}{\text{Run}}=(1)/(4)`

y-intercept = -4

Equation → y =
`(1)/(4)x-4`