How do you write an equation in slope-intercept form for a line containing #(9,-3)# and #(5, 5)#?

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Answer 1 · 2017-06-07T02:33:04

#y=-2x+15#

The general form of a linear equation is #y=mx+c# where #m# is the gradient (slope) and #(0,c)# is the #y#-intercept.

#m=(y_2-y_1)/(x_2-x_1)# for a line containing #(x_1,y_1)# and #(x_2,y_2)#.

For #(9,-3)# and #(5,5)#,
#m=(5-(-3))/(5-9)#
#m=-2#

Therefore, #y=-2x+c#. To find c, substitute #(5,5)# into this equation.
#5=-2(5)+c#
#c=5+10#
#c=15#

Substitute #c# value back into equation,
#y=-2x+15#

Answer 2 · 2017-06-07T02:41:28

#y=-2x+15#

We first begin with finding out the slope using the slope formula, which

is #(y_2-y_1)/(x_2-x_1)#. Plug in our points #(5-(-3))/(5-9)#=#-8/2#=#-2#. This is our slope now we use the point slope formula #y-y_1=m(x-x_1)#.

You can pick any of the two points, #m# is our slope which is #-2#.

#y-5=-2(x-5)# go ahead and solve.

You should arrive to the answer of #y=-2x+15#.

When you do #(y_2-y_1)/(x_2-x_1)# it doesn't matter which point is your #(y_2,x_2)# or #(y_1,x_1)#.