To find the equations of the parties of a triangle, first of all it is necessary to try to resolve an issue of how to find the straight line equation on the plane if its directing s vector (m, n) and some point of M0 (x0, y0) belonging to a straight line is known.
Instruction
1. Take any (variable, floating) the M point (x, y) also construct M0M vector = {x-x0, y-y0} (it is possible to write down also M0M (x-x0, y-y0)) which, obviously kollinearen will (be parallel) in relation to s. Then, it is possible to conclude that coordinates of these vectors are proportional therefore it is possible to work out the initial equation of a straight line: (x-x0) / m = (y-y0)/n. This ratio will be used further at the solution of an objective.
2. All further actions are defined proceeding from a way of a task of a triangle. the 1st way. The triangle is set by coordinates of points of three of its tops that in school geometry corresponds to the setting lengths of three of its parties (see fig. 1). That is in a condition M1 points (x1, y1) are given, to M2 (x2, y2), by M3 (x3, y3). Correspond to them their radius vectors) OM1, 0M2 and OM3 with same, as well as at points, coordinates. Receiving the equation of the party of M1M2 requires its directing M1M2=OM2 vector – OM 1= M1M2 (x2-x1, y2-y1) and any of points of M1 or M2 (the point with the smaller index is taken here).
3. So, for the party of M1M2 the initial equation of a straight line (x-x1) / (x2-x1)= (y-y1)/(y2-y1). Working purely inductively it is possible to write down the equations of other parties. For the party of M2M3: (x-x2) / (x3-x2)= (y-y2)/(y3-y2). For the party of M1M3: (x-x1) / (x3-x1)= (y-y1)/(y3-y1).
4. the 2nd way. The triangle is set by two points (the same, as before M1 (x1, y1) and M2 (x2, y2)) and also Horta of the directions of two other parties. For the party of M2M3: p^0 (m1, n1). For M1M3: q^0 (m2, n2). Therefore the answer for the party of M1M2 will be the same, as in the first way: (x-x1) / (x2-x1)= (y-y1)/(y2-y1).
5. For the party of M2M3 as a point (x0, y0) the initial equation undertakes (x1, y1), and the directing vector is p^0 (m1, n1). For the party of M1M3 as a point (x0, y0) the directing vector – q^0 undertakes (x2, y2) (m2, n2). Thus, for M2M3: equation (x-x1) / m1=(y-y1)/n1. For M1M3: (x-x2) / m2=(y-y2)/n2.