How to draw a curve without knowing 4 points?

According to cairo example code, following code

double x=25.6,  y=128.0;
double x1=102.4, y1=230.4,
       x2=153.6, y2=25.6,
       x3=230.4, y3=128.0;

cairo_move_to (cr, x, y);
cairo_curve_to (cr, x1, y1, x2, y2, x3, y3);

cairo_set_line_width (cr, 10.0);
cairo_stroke (cr);

cairo_set_source_rgba (cr, 1, 0.2, 0.2, 0.6);
cairo_set_line_width (cr, 6.0);
cairo_move_to (cr,x,y);   cairo_line_to (cr,x1,y1);
cairo_move_to (cr,x2,y2); cairo_line_to (cr,x3,y3);
cairo_stroke (cr);

can generate the curve and two pink lines.

But that need 4 points, (x,y), (x1,y1), (x2,y2), (x3,y3)

If I only have x,y and x3, y3 (start and end points of the curve), is there any math fo开发者_JAVA百科rmula to generate those pink lines without knowing x1,y1 and x2,y2?

Edit:

Its for the case that I generate the curve through following way.

cairo_move_to (cr, x, y);
cairo_curve_to (cr, x, y3, x3, y, x3, y3);

Just make the points up:

• start with your two known points, (x₁,y₁) and (x₃,y₃):

• join the two lines:

• create P₂ as halfway between P₁ and P₃:

• now rotate P₃ 90° clockwise:

• do with same with P₄, create it halfway between P₁ and P₃:

• rotate P₄ 90° clockwise:

• Now you have your four points, and can draw your bézier curve:

The midpoint can be calculated as:

P_mid = (x₁+x₃)/2 , (y₁+y₃)/2

double x1=25.6,  y1=128.0;
double x3=153.6, y3=25.6;

double xm = (x1+x3)/2;
double ym = (y1+y3)/2;

//rotate Pm by 90degrees around p1 to get p2
double x2 = -(ym-y1) + y1;
double y2 =  (xm-x1) + x1;

//rotate Pm by 90degrees around p3 to get p4
double x4 = -(ym-y3) + y3;
double y4 =  (xm-x3) + x3;

Not unless you provide some kind of constraint that can be used to derive the location of the pink lines. By themselves, two end-points can only define a straight line segment.

The pink lines represent the departure vectors from the two end points. Without these vectors, the "curve" between the two points is simply a straight line (unless you have some other information that defines it).

If you don't have (x1,y1) and (x2,y2), you can just use (x3,y3) as the endpoint for the pink line from (x,y) and vice versa. They'll end up right on top of your black line, which is where they should be for a straight line.

If the curve is defined by a function, calculate the derivative as you approach the end points and draw the tangent line along that angle.