How do you write an equation in slope intercept form given m=1/3, and contains (-3,-15)?

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Answer 1 · 2018-04-03T06:44:37

-14 .Check below

Use the equation
#y-y_1 =m(x-x_1)#

Where,

#y_1 = -15#
#x_1 = -3#
#m =1/3#

Next the equation is :
#y+15=1/3(x+3)#
#3y+45=x+3#
Thus,
#y=(x-42)/3#
or u can write
#y=1/3 x - 14#

Now u can use the equation of straight line which is
#Y=mX+c#

compare the above equation with your result and the y-intercept is -14

Answer 2 · 2018-04-03T06:52:49

#y =1/3x -14#

#m # is the slope/gradient of the line

#y =mx + c# is the general equation of the straight line

So we know #y =1/3x +c# and to find the value of #c# we substitute the numbers in the coordinate that the line passes through

#y =1/3x + c# (-3,-15)

#-15 =1/3 ##xx##-3 + c#

#-15 =-1 +c#

#c = -14#

So the equation is # y =1/3x -14#