What is the equation of the line between #(0,0)# and #(25,-10)#?

Slope

First determine the slope using the formula:

#m=(y_2-y_1)/(x_2-x_1),#

where:

#m# is the slope, #(x_1,y_1)# is one point, and #(x_2,y_2)# is the second point.

Plug in the known data. I am going to use #(0,0)# as the first point, and #(25,-10)# as the second point. You can do the opposite; the slope will be the same either way.

#m=(-10-0)/(25-0)#

Simplify.

#m=-10/25#

Reduce by dividing the numerator and denominator by #5#.

#m=-(10-:5)/(25-:5)#

#m=-2/5#

The slope is #-2/5#.

Point-slope form

The formula for the point-slope form of a line is:

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

where:

#m# is the slope, and #(x_1,y_1)# is the point. You can use either point from the given information. I'm going to use #(0,0)#. Again, you can use the other point. It will end up the same, but take more steps.

#y-0=-2/5(x-0)# #larr# point-slope form

Slope-intercept form

Now we can determine the slope-intercept form:

#y=mx+b,#

where:

#m# is the slope, and #b# is the y-intercept.

Solve the point-slope form for #y#.

#y-0=-2/5(x-0)#

#y=-2/5x# #larr# slope-intercept form #(b=0)#

Standard form

We can convert the slope-intercept form into the standard form for a linear equation:

#Ax+By=C,#

where:

#A# and #B# are integers, and #C# is the constant (y-intercept)#

#y=-2/5x#

Eliminate the fraction by multiplying both sides by #5#.

#5y=(-2x)/color(red)cancel(color(black)(5))^1(color(red)cancel(color(black)(5)))^1#

#5y=-2x#

Add #2x# to both sides.

#2x+5y=0# #larr# standard form

graph{y=-2/5x [-10, 10, -5, 5]}