On the basis of the axiom describing properties of a straight line: what was a straight line, there are points which are belonging and not belonging to it. Therefore it is quite logical that not all points will lie on one straight line.
It is required to you
- - pencil;
- - ruler;
- - handle;
- - notebook;
- - calculator.
Instruction
1. It is quite simple to check accessory of a point of that or other straight line. Use for this purpose the straight line equation. So, suppose, that the straight line passes through points And (x1, y1) and In (x2, y2). The point is given To (x, y): it is necessary to check its accessory of a straight line. The line equation on two points has the following appearance: (x - x1) * (y2 - y1) - (x2 - x1) * (y - y1) = 0.
2. Substitute value of coordinates of a point To in the equation. If (x - (y2 - y1) - (x2 - x1) * (y - y1) more than zero is x1) *, then the point To is located more to the right of or below the straight line drawn on points And yes Century.
3. In case (x - (y2 - y1) - (x2 - x1) * (y - y1) less than zero is x1) *, the point To is located above or lines are more left. In other words, only if the look equation (x - x1) * (y2 - y1) - (x2 - x1) * (y - y1) = 0 is fair, points And, In and To will be located on one straight line.
4. In other cases only two points (And yes C) which, on a task condition, lie on a straight line, will belong to it: (a point J) the straight line will not pass through the third point.
5. Consider the second option of definition of accessory of a point by the prima: this time it is necessary to check whether the point With (x, y) belongs to a piece with trailer points In (x1, y1) and And (x2, y2) which is a part of direct z.
6. Describe points of the considered piece the equation of pOB+(1-p) OA=z provided that 0≤p≤1. OV and OA are vectors. If there is such number p which is more or equally 0, but it is less or equally 1, then pOB+(1-p) OA=C, so, a point With will lie on AV piece. Otherwise, this point will not belong to this piece.
7. Paint equality of pOB+(1-p) OA=C coordinate-wise: px1+(1-p)x2=x and py1+(1-p)y2=y.
8. Find from the first equation number r and substitute its value in the second equality. If equality corresponds to conditions 0≤p≤1, then the point With belongs to AV piece.
9. Construct points on the set coordinates and draw through them a straight line. It will allow to see the points lying on one straight line and those points that do not belong to it.