Question

What is the equation in standard form of the line shown on the graph

Answer 1

Explanation:

We take two points (0,-3) and (4,0)

eq. of line thro' two points is y-y1=m(x-x1)

m=(y2-y1)/(x2-x1)=(0+3)/(4-0)=3/4

Reqd. eq . is y-(-3)=3/4(x-0)

or y=3/4(x)-3

Answer 2

For this case we have that the standard equation of a line is given by:

`y = mx + b`

Where:

m: It's the slope

b: It is the cut point with the y axis.

`m = \frac {y2-y1} {x2-x1}`

We have the following points of the graph:

`(x1, y1) = (4,0)\\(x2, y2) = (0, -3)`

Substituting:

`m = \frac {-3-0} {0-4}\\m = \frac {-3} {- 4}\\m = \frac {3} {4}`

We have:

`y = \frac {3} {4} x + b`

To find the cut point we substitute a point:

`0 = \frac {3} {4} (4) + b\\0 = 3 + b\\b = -3`

Thus, the equation is:

`y = \frac {3} {4} x-3`

Answer:

`y = \frac {3} {4} x-3`