|p| = 2.
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P2 + 13p - 30 = 0 Answer: p= -15, p = 2
It depends on what aspect of constant velocity you are talking about. Since the velocity is not changing, one valid equation is: V = [number] At the same time, acceleration is zero, so another equation is: A = 0 If "p" is position and p1 is the original position and p2 is the current position after tine lapse "t," then: p2 = p1 + Vt
p2 + 2pq + q2 = 1q2 + 2pq + (p2 - 1) = 0q = 1/2 [ -2p plus or minus sqrt( 4p2 - 4p2 + 4 ) ]q = -1 - pq = 1 - p
You use the well-known point-point formula . Here is it: suppose P1 = (x1,y1) and P2 = (x2,y2) An equation for the line containing P1 and P2 is y - y1 = [(y1-y2)/(x1-x2)] (x -x1) Note that the quantity in brackets is the slope of the line. Note also that it does not matter which point is P1 and which is P2.
An equation that is always true is an identity.