Question

TY12What is the equation of the line that is parallel to thegiven line and passes through the point (12,-2)?10.8(126)6-42O y=-x + 10O y=-x + 12O y=-x-10O y = x - 12 A-10 -8 -6 -4 -22610 12 14 X12.-2(0.44)181012

Answer 1

Let's find the slope of the given line:

`\begin{gathered} \text{Let:} \\ (x1,y1)=(0,-4) \\ (x2,y2)=(12,6) \\ m=(y2-y1)/(x2-x1)=(6-(-4))/(12-0)=(10)/(12)=(5)/(6) \end{gathered}`

Since the lines are parallel, we can conclude:

`m1=m2`

Using the point-slope equation:

`\begin{gathered} y-y1=m(x-x1) \\ y-(-2)=(5)/(6)(x-12) \\ y+2=(5)/(6)x-10 \\ y=(5)/(6)x-10-2 \\ y=(5)/(6)x-12 \end{gathered}`