# How do you find slope, point slope, slope intercept, standard form, domain and range of a line for Line I (3,4) (0,10)?

##### 1 Answer

Slope:

Point-Slope Equation:

Slope-Intercept Equation:

Standard Form Equation:

Domain:

Range:

#### Explanation:

Let's start with domain and range. Think about drawing a straight line on a coordinate plane. As long as the line is not horizontal or vertical, you should be able to choose any

In simpler terms, no value of

#x# will make#y# undefined for a linear equation. So, the domain is all real numbers, or#x in RR# .In simpler terms, any non-horizontal line stretches from the very bottom of the Cartesian plane (

#-oo# ) to the very top (#oo# ), so the range (or all possible#y# values) is all real numbers, or#y in RR# .

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Now, let's find the slope of our line. Everything else will follow:

#"slope" = "change in y"/"change in x" = (10-4)/(0-3) = 6/-3 = -2#

The point-slope equation for a line looks like this:

#y - y_1 = m(x-x_1)# Where

#(x_1,y_1)# is a point on the line and#m# is the slope.Using

#(0,10)# as our point and#-2# as our slope, the equation is:

#y - 10 = -2(x-0)# Or:

#y - 10 = -2x#

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Now to find slope-intercept. We are looking for something that looks like this:

#y = mx+b# Where

#m# is the slope and#(0,b)# is the#y# intercept. Luckily, one of the points we were given (0,10) IS the#y# intercept, since its#x# value is 0. Therefore,#b=10# , and we already know#m=-2# . So:

#y = -2x+10#

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Finally, we need to find the standard form equation. It will look like this:

#ax+by=c# Where

#a# ,#b# , and#c# are integers with a GCF of 1, and#a# is positive.To find the standard form, let's take our slope-intercept form and manipulate it until we get something that looks like standard form:

#y = -2x+10#

#y+2x = -2x+10+2x#

#2x + y = 10#

This meets all the requirements for standard form, so it is our final equation!

*Final Answer*