function(a){if(this===void 0||this===null)throw new TypeError;var
d=Object(this),c=d.length>>>0;if(c===0)return-1;var
b=0;arguments.length>0&&(b=Number(arguments[1]),b!==b?b=0:b!==0&&b!==1/0&&b
!==-(1/0)&&(b=(b>0||-1)*Math.floor(Math.abs(b))));if(b>=c)return-1;for(b=b>=
0?b:Math.max(c-Math.abs(b),0);b<c;b++)if(b in d&&d[b]===a)return b;return-1}
P jest środkiem odcinków AC i BD
| xC+xA | yC+yA | |||
zatem xP= | i yP= | |||
| 2 | 2 |
| xB+xD | yB+yD | |||
oraz xP= | i yP= | |||
| 2 | 2 |
function(a){if(this===void 0||this===null)throw new TypeError;var
d=Object(this),c=d.length>>>0;if(c===0)return-1;var
b=0;arguments.length>0&&(b=Number(arguments[1]),b!==b?b=0:b!==0&&b!==1/0&&b
!==-(1/0)&&(b=(b>0||-1)*Math.floor(Math.abs(b))));if(b>=c)return-1;for(b=b>=
0?b:Math.max(c-Math.abs(b),0);b<c;b++)if(b in d&&d[b]===a)return b;return-1}