Find the equetion of the line that passes through the pont (1,-6) and is parallel to the line x+2y=6.

PLZ SHOW THE STEPS

x+2y=6
2y=6-x
y=-x/2+3

m=-1/2
Point slope form: y-y1=m(x-x1)
Plug it in.
y+6=-1/2(x-1)
Distribute and simplify.
y+6=-x/2+1/2
y=-x/2-11/2

So the answer is -x/2-11/2.

If you have learned how to take the derivative, this way could be useful. : )

*Note: When asked to find the equation of a line that is parallel to another line, you must remember that both of the lines slopes (m) will be the same. ---> m1 = m2 ---> Slope Line 1 = Slope Line 2

Now the actual work.

1) Solve for y

---> x + 2y = 6

---> y = (6 - x) / 2

---> y = 3 - (x / 2)

2) Find the slope by taking the derivative of y

---> y' = ( 3 - (x / 2) )'

---> y' = - ( ( x )' * ( 2 ) - ( x ) * ( 2 )' ) / ( 2 )^2

---> y' = - ( ( 1 )*( 2 ) - ( x )*( 0 ) ) / ( 4 )

---> y' = - 2 / 4 ------> - 1 / 2

3) Now finally use the y - y1 = m * ( x - x1 ) ---> Point-Slope Form

Plug in your points to their " x,y " variables

---> y = ( -6 ) - ( 1 / 2 ) * ( x - 1 )

---> y = -6 - ( 1 / 2 )x + ( 1 / 2 )

---> y = - ( 1 / 2 )x - ( 11 / 2 )

find the slope of x+2y=6
y= -1/2x + 3
-1/2 is the slope that you need with the point (1,-6)
point-slope form:
y+6=-1/2(x-1)
y+6=-1/2x + 1/2
-6 -6
y = -1/2x - 5 1/2
Answer:
y = -1/2x - 5 1/2

2y=6-x

Y=3-1/2x

-6=-1/2*1+b

B. -51/2

Y=-1/2x-51/2