Write the standard form of the equation of each line passing through P with slope m. P(subscript 1)=(4,-5); P(subscript 2) = (11,2)
\[P _{1} = (4, -5); P _{2} =(11,2)\]
move it all by placing P1 at the origin ....
move a points by doing the same thing to each one 4, -5 11, 2 -4 +5 -4 +5 --------------- 0 0 7 7 so the line is parallel to the this one thru the origin
<y> = <x> , such that <y> = y-5, and <x> = x+5 to account for moving all the points of the original line by the same amount
then just put it into standard form: Ax + By = C
@amistre64 I am totally new to this stuff so can you like be as simple as possible? thanks
well the answer amistre is right... the formula is correct and gt the same answer as him
moving points is the simplest way to go. its either that or you can try to remember a formula that just tells you to do the same thing
spose you have 2 points: (x1,y1) and (x2, y2) ; move the points by x1,y1 (x1, y1) and (x2, y2) -x1 -y1 -x1 -y1 -------------------- 0 0 x2-x1, y2-y1 , define the slope by y/x \[slope=\frac{y_2-y_1}{x_2-x_1}\] then the point slope formula is created as: (y-y1) = slope(x-x1), which is simply subbing in the movements for <y> and <x>
since the slope is 7/7 = 1 <y> = 1<x>, or (y-4) = (x+5) y-4 = x+5 -x + y -4 = 5 -x + y = 9 and since they dont like a negative x, multiply thru by -1 x - y = -9
since i tend to mess up trying to put values into a formula; and if you dont put them inthe right place you tend to mess up the whole thing ... i just work it the sensible way :)
well theres a lil prob with the last part on the equation u jus did
lets test it then x - y = -9 4 -5 x - y = -9 11 2 yeah, i misplaced a negative along the way
(x1, y1) and (x2, 2) -x1 -y1 -x1 -y1 -------------------- 0 0 ok i understand this part but on the other side i am confused on (that right side)
whats ur point of confusion???
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