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Mathematics 11 Online
OpenStudy (karatechopper):

write equation in standard form for a line passing through (1,6) with a slope of 2

OpenStudy (amistre64):

standard form is in "standard form" or as in slope intercept form which tends to be used as a standard form for the equation of a line?

OpenStudy (amistre64):

standard form: ax +by = c, where slope = -a/b 2 -a -- = --- ; therefore a = -2, and b=1 1 b -2x +b = c ; use the given point to calibrate for c -2(1) + 6 = c -2+6 = c 4 = c ; rewrite -2x +b = 4 ; the only issue with this is that some people hate to see a negative in front so just multiply it all by -1 to get: 2x -b = -4

OpenStudy (karatechopper):

?

OpenStudy (amistre64):

if "slope intercept" is their definition for standard form; then we can go this route: y = mx -mPx +Py ; such that m=slope (2) and P(x,y) = Px,Py (1,6) y = 2x -2(1) + 6 y = 2x -2+6 y = 2x +4

OpenStudy (amistre64):

that first is typoed; should read: 2x -y = -4 :)

OpenStudy (karatechopper):

sorry i do not understand u

OpenStudy (amistre64):

well, what do you know about line equation forms?

OpenStudy (karatechopper):

how about i just say i started out with slope intercept form, y=mx+b, then i plugged in the numbers, 6=2(1) +b, then i simplified, 6=2+b, then i figured out b=4. and now i stuck

OpenStudy (amistre64):

you did good so far; now re write your y=mx+b with all the information you found plugged in y = mx + b y = 2x + 4 right?

OpenStudy (karatechopper):

correct, but should i not plug in the x's and y's? why?

OpenStudy (amistre64):

the x and y parts are generic; they represent a point that you have not found and will be found by actually using the equation.... we need them to be filler in order to find all points on the line.

OpenStudy (amistre64):

we can dbl check that the given point is ON our line by plugging in the x and y values into our generic spots for x and y

OpenStudy (amistre64):

is the point (1,6) on our line? y = 2x + 4 6 = 2(1) + 4 6 = 6 ...... yes it is ...................................... is the point (4,4) on our line? lets check and see y = 2x + 4 4 = 2(4) + 4 4 = 12 .... this is a false statement so we know (4,4) is not on our line

OpenStudy (amistre64):

we need that y and x to be left generic in order to find all the points that fit our line

OpenStudy (karatechopper):

got it but what would the equation be? 2x+4=y?

OpenStudy (amistre64):

well, that is a fine looking equation; but by convention; that is not the results for this problem :) standard form is defined by the format: ax+by=c ; where a is a positive number lets take the equation you found y = 2x + 4 ; and lets see if we can re shape it into ax +by = c; we can subtract 2x from each side to get: -2x +y = 4 ; looks better, but its got a negative in front; let multiply it all by -1 to get: 2x -y = -4 ; this fits our format for ax +by = c

OpenStudy (karatechopper):

ok thank u!

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