How do you write an equation of a line given (9,-5), m=-1/3?

1 Answer
Aug 29, 2017

#y=-1/3x-2#

Explanation:

Let the given point be #P_1->(color(green)(x_1),color(red)(y_1))=(color(green)(9),color(red)(-5))#

Let the gradient be #m=-1/3#

Using the standardised form #color(red)(y_1)=mcolor(green)(x_1)+c#

Where #c# is a constant

Then by substitution: #color(red)(-5)=(-1/3)(color(green)(9))+c#

but #-1/3xx9 = -3# giving:

#-5=3+c#

Add #color(magenta)(3)# to both sides

#color(green)(-5=-3+c color(white)("dd")->color(white)("dd")-5color(magenta)(+3)=-3color(magenta)(+3)+c)#

#color(green)(color(white)("ddddddddddddd")-> color(white)("dddd")-2color(white)("d")=color(white)("dddd")0color(white)("d")+c#

Thus #c=-2# so the finished equation is:

#y=-1/3x-2#