How do you write the equation given (-6,7); parallel to 3x + 7y = 3?

2 Answers

Answer:

#s: y = -3/7 x + 31/7#

Explanation:

We take the line above in the form #r:y = ax + b#
We know that any parallel is #s:y = ax + c#
We choose #(-6, 7) in s#

#7y = 3 - 3x => r: y=-3/7x+3/7#

#s:7 = -3/7 (-6) + c#

#49 = 18 + 7c => c = 31/7#

Jul 11, 2017

Answer:

#y=-3/7x+31/7#

         or

#7y=-3x+31#

Explanation:

Change #3x+7y=3# to standard form of #y=mx+c#
#7y=3-3x#
#y=-3/7x+3/7#

gradient, #m#, can be determined as #-3/7#

Two parallel lines would have the same gradient, in this case gradient of #-3/7#

You can choose to use gradient formula
Gradient, #m=(y_1-y_2)/(x_1-x_2)#
or general formula for straight line
#(y=mx+c)#

I would first be attempting it using gradient formula

replace #m# with #-3/7#; replaace #x_1,x_2,y_1,y_2# with #x# and the x-coordinate, #y# and the y-coordinate respectively

in this case, you have only 1 point given, if there is more, the x and y coordinate must be from the same point.

#-3/7=(y-7)/(x-(-6))#
#-3/7=(y-7)/(x+6)#

rearrange
#-3/7(x+6)=(y-7)/(x+6)#
#-3/7x-18/7=y-7#
#y=-3/7x-18/7+7#
#y=-3/7x+31/7#

in case you don't like fractions, multiply whole equation by 7
#7y=-3x+31#

Using general formula

replace #m# with #-3/7#; #x# with the x-coordinate, #y# with the y-coordinate

in this case, you have only 1 point given, if there is more, the x and y coordinate must be from the same point.

#7=(-3/7)(-6)+c#

solve for c
#7=18/7+c#
#c=31/7#

substitute #m# and #c# into #y=mx+c#
#y=-3/7x+31/7#
#7y=-3x+31#

Both equation for straight line you get are the same, depending on which you prefer.