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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

TY12What Is The Equation Of The Line That Is Parallel To Thegiven Line And Passes Through The Point 122108126642O Yx 10O Yx 12O Yx10O Y X 12 A10 8 6 4 22610 12 class=

Sagot :

Let's find the slope of the given line:

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

Since the lines are parallel, we can conclude:

[tex]m1=m2[/tex]

Using the point-slope equation:

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

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