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

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 #x# or #y# value and find a point on the line with that value.

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