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line AB contains points (2,1) and (-1,-8). what is the equation of the line parallel to line AB that contains point (0,2)?

Sagot :

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[tex]General\ equation\ for\ line\ in\ slope\ intercept\ form:\\\\y=ax+b\\\\To\ find\ a\ and\ b\ substitude\ points\ (2,1),\ (-1,-8)\ into\ equation\\\\ \left \{ {{1=2a+b}\ \ \ \ \atop {-8=-1a+b\ |*-1}} \right.\\\\ \left \{ {{1=2a+b}\ \ \ \ \atop {8=1a-b}} \right.\\\\ +----\\\\Addition\ method\\\\ 9=3a\ \ |:3\\\\ a=3\\\\ b=1-2a=1-6=-5\\\\ AB:\ y=3x-5\\\\ new\ line: \ y=mx+c\\ if\ line\ is\ parallel\ m=3\\\\ y=3x+c\ \ \ | subtitute\ (0,2)\\\\ 2=c\\\\ Answer: y=3x+2[/tex]

Answer:

Step-by-step explanation:

(2, 1 ) and (-1, -8) are on the line AB

Find the slope of the line by substituting the given values in the slope formula:

m= (y2 - y1)/(x2 - x1)

m= (-8 - 1 )/(-1 - 2)

m= 3

By Point slope form:

y - y1 = m ( x - x1)

y - 1 = 3 ( x - 2)

y = 3x - 5 This is the equation of the line AB.

The line parallel to y= 3x - 5 that pases through ( 0, 2 ).

Parallel lines have equal slope, so the slope for both lines is 3.

Now, we have:

Using point slope form ( 0, 2 )

y - 2 = 3 (x - 0)

y= 3x + 2

The equation of the line that is parallel to line AB is : y = 3x + 2.

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