The straight line on the plane is unambiguously set by two points of this plane. Between two straight lines understand length of the shortest piece as distances between them, that is length of their general perpendicular. The shortest joint perpendicular for two set straight lines is a constant. So that to answer a question of an objective, it must be kept in mind that the distance between two set parallel straight lines is found is on the set plane. It would seem that there is nothing more simply: to take any point on the first straight line and to lower from it a perpendicular on the second. Compasses and a ruler it is elementary to make it. However it is only an illustration of the forthcoming decision which means exact calculation of length of such joint perpendicular.
It is required to you
- - handle;
- - paper.
Instruction
1. For the solution of an objective it is necessary to use methods of analytical geometry, having attached the plane and straight lines to the system of coordinates that will allow not only to calculate precisely necessary distance, but also to leave from the explaining illustrations. The main equations of a straight line on the plane have the following look.1. Straight line equation, as linear function graph: y=kx+b.2. General equation: Ax+By+D=0 (here n= {A, B } is a vector of a normal to this straight line).3. Initial equation: (x-x0) / m = (y-y0)/n. Here (x0, yo) – any point lying on a straight line; {m, n } =s are coordinates of its directing s vector. It is obvious that if there is a search of the perpendicular straight line set by the general equation, then s=n.
2. Let the first of parallel direct f1 be set by y=kx+b1 equation. Having transferred expression to a general view, at you kx-y+b1=0, that is A=k, B=-1 will turn out. N= {k,-1 } will be a normal to it. Now it is necessary to take any abscissa of a point h1 on f1. Then its ordinate of y1=kx1+b1. Let the equation of the second of parallel direct f2 will have an appearance: at =kx+b2 (1) where k is identical to both straight lines, owing to their parallelism.
3. Further you need to work out the initial equation of the line perpendicular to both f2, and f1 containing the M point (x1, y1). At the same time believe that h0= h1, y0=y1, S= {k,-1 }. As a result at you has to the following equality will turn out: (x-x1) / k = (y-kx1-b1)/(-1) (2).
4. Having solved the system of the equations consisting of expressions (1) and (2) you will find the second point defining required distance between parallel direct N (x2, y2). The required distance will be equal to d= | to MN |= (x2-x1) ^2+(y2-y1) ^2) ^1/2.
5. Example. Let the equations of the set parallel straight lines on f1 plane – at =2x +1 (1); f2 – y=2x+5 (2). We take any point h1=1 on f1. Then y1=3. The first point, will have thus coordinates of M (1.3). Equation of the general perpendicular (3): (x-1)/2 = - y+3 or y=-(1/2)x+5/2. Having substituted this value y in (1), it is possible to receive: - (1/2) x+5/2=2kh +5, (5/2) x =-5/2, h2=-1, y2=-(1/2)(-1) +5/2=3. The second basis of a perpendicular is in a point with coordinates N (-1, 3). The distance between parallel straight lines will make: d= | MN |= ((3-1) ^2+ (3+1) ^2) ^1/2= (4+16) ^1/2=4.47.