What is the equations of a line that is parallel to y = -3x + 6 and contains the point (-3,-5)?

2 Answers
Jun 21, 2018

Answer:

#y=-3x-14#

Explanation:

Any line parallel to #y=-3x+6#
will be in the form #y=-3x+c# for some constant #c#

If #(x,y)=(-3,-5)# is a solution to such an equation, then replacing #y# with #(-5)# and #x# with #(-3)# gives
#-5=(-3) *(-3)+c#

#rarr -5=9+c#

#rarr c=-14#

So the desired parallel line would have the equation
#y=-3x-14#

Jun 21, 2018

Answer:

The equation of the line parallel to #y=-3x+6# that passes through the point #(-3,-5)# is:

#y=-3x-14#

Explanation:

The line #y=-3x+6# is already in 'slope-intercept' form, so we can just read off the slope (gradient) as being #-3#.

We want a line with a gradient of #-3# (because that's what 'parallel' means - having the same gradient) that passes through the point #(-3,-5)#.

Take the form #y=mx+c# and substitute in the 3 things we already know: the slope, one value of #x# and one value of #y#:

#-5=(-3)(-3)+c#

#c# is the y-intercept of the line: the #y# coordinate of the point at which it cuts the #y# axis (which is the line #x=0#).

#-5=9+c#

Subtract #9# from both sides:

#-14=c#

That means the equation of the line is #y=-3x-14#