What is the equation of the line passing through #(13,-4)# and #(14,-9)#?

I'm not sure which form of equation you want it to be in, but going to show the simplest, or point-slope form, which is #y - y_1 = m(x-x_1)#.

First, we need to find the slope of the line, #m#.

To find the slope, we use the formula #m = (y_2-y_1)/(x_2-x_1)#, also known as "rise over run", or change of #y# over change of #x#.

Our two coordinates are #(13, -4)# and #(14, -9)#. So let's plug those values into the slope equation and solve:
#m = (-9-(-4))/(14-13)#
#m = -5/1#
#m = -5#

Now, we need a set of coordinates from the given or the graph. Let's use the point #(13, -4)#

So our equation is:
#y-(-4) = -5(x-13)#

Simplified...
#y + 4 = -5(x-13)#

#"the equation of a line in "color(blue)"slope-intercept form"# is.

#•color(white)(x)y=mx+b#

#"where m is the slope and b the y-intercept"#

#"to calculate m use the "color(blue)"gradient formula"#

#color(red)(bar(ul(|color(white)(2/2)color(black)(m=(y_1-y_1)/(x_2-x_1))color(white)(2/2)|)))#

#"let "(x_1,y_1)=(13,-4)" and "(x_2,y_2)=(14-9)#

#rArrm=(-9-(-4))/(14-13)=-5#

#rArry=-5x+blarrcolor(blue)"is the partial equation"#

#"to find b use either of the two given points"#

#"using "(13,-4)#

#-4=-65+brArrb=61#

#rArry=-5x+61larrcolor(red)"in slope-intercept form"#