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Mathematics
OpenStudy (anonymous):

how do you write an equation in point-slope form with the slope of 3 and points (2,5) *really important question PLEASE answer it

OpenStudy (amistre64):

point slope form, not slope intercept form

OpenStudy (amistre64):

y-Py = m(x-Px)

OpenStudy (amistre64):

they give you the slope and the (Px,Py)

OpenStudy (saifoo.khan):

:O My bad. :/

OpenStudy (amistre64):

'sok, they all look alike after wahile lol

OpenStudy (saifoo.khan):

y = mx + b 5 = 3(2) + b 5 = 6 + b b = 5-6 b = -1 (0,-1) Now, \[(y_2-y_1)= m(x_2-x_1)\]

OpenStudy (saifoo.khan):

\[y-5 = 3 (x-2) \]

OpenStudy (amistre64):

the equation of a line start by finding a slope, and all the other forms are but an algebraic manipulation of that: given 2 points: (Px,Py) and (x,y) , slope is defined as the difference in the y values with respect to the difference in the x values: so subtract the points and divide y by x (x , y) -(Px,Py) ------- (x-Px,y-Py); slope, or m as it is so named, = y/x: y-Py ---- = m ; from this we can get the "point slope" form x-Px y-Py = m(x-Px) ; and from this we can obtain the "slope intercept" form y = mx + [-mPx+Py] y = mx + c; if we split the slope back up into a rational form we can manipulate it into the "standard" form for the equation of a line: y = (-a/b)x + c ; such that c is any constant, even c/b would be suitable y = (-a/b)x + c/b by = -ax + c ax + by = c ; and from this we can get the "intercept form" of the line ax by c -- + -- = -- c c c x y --- + --- = 1 ; where c/a and c/b are the intercepts c/a c/b

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