假设你是前端工程师，在canvas上已知道起点和终点对象的坐标，起点和终点的坐标是一个object里面有canvasX和canvasY两个值分别代表画布上的xy值，想要绘制从起点指向终点的连线，并且在终点方向有起点到终点距离1/6大小的三角形箭头

好的，可以使用canvas的API来实现这个需求。

首先，需要获取canvas对象以及绘图环境context：

let canvas = document.getElementById('myCanvas');
let ctx = canvas.getContext('2d');

然后，根据起点和终点的坐标，绘制一条连线：

ctx.beginPath();
ctx.moveTo(start.canvasX, start.canvasY);
ctx.lineTo(end.canvasX, end.canvasY);
ctx.stroke();

接着，需要计算出箭头的位置和朝向。可以使用向量的方式来计算。将起点和终点的坐标看成向量的起点和终点，箭头位置就是终点向起点方向的向量，长度为起点到终点距离的1/6。箭头朝向就是这个向量的方向。

let dx = end.canvasX - start.canvasX;
let dy = end.canvasY - start.canvasY;
let length = Math.sqrt(dx * dx + dy * dy);
let arrowLength = length / 6;
let arrowDirection = Math.atan2(dy, dx);
let arrowX = end.canvasX - arrowLength * Math.cos(arrowDirection);
let arrowY = end.canvasY - arrowLength * Math.sin(arrowDirection);

最后，在箭头位置绘制一个三角形即可：

ctx.beginPath();
ctx.moveTo(arrowX, arrowY);
ctx.lineTo(arrowX + arrowLength * Math.cos(arrowDirection - Math.PI / 6), arrowY + arrowLength * Math.sin(arrowDirection - Math.PI / 6));
ctx.lineTo(arrowX + arrowLength * Math.cos(arrowDirection + Math.PI / 6), arrowY + arrowLength * Math.sin(arrowDirection + Math.PI / 6));
ctx.closePath();
ctx.fill();

完整代码如下：

let canvas = document.getElementById('myCanvas');
let ctx = canvas.getContext('2d');

let start = { canvasX: 50, canvasY: 50 };
let end = { canvasX: 150, canvasY: 150 };

// 绘制连线
ctx.beginPath();
ctx.moveTo(start.canvasX, start.canvasY);
ctx.lineTo(end.canvasX, end.canvasY);
ctx.stroke();

// 计算箭头位置和朝向
let dx = end.canvasX - start.canvasX;
let dy = end.canvasY - start.canvasY;
let length = Math.sqrt(dx * dx + dy * dy);
let arrowLength = length / 6;
let arrowDirection = Math.atan2(dy, dx);
let arrowX = end.canvasX - arrowLength * Math.cos(arrowDirection);
let arrowY = end.canvasY - arrowLength * Math.sin(arrowDirection);

// 绘制箭头
ctx.beginPath();
ctx.moveTo(arrowX, arrowY);
ctx.lineTo(arrowX + arrowLength * Math.cos(arrowDirection - Math.PI / 6), arrowY + arrowLength * Math.sin(arrowDirection - Math.PI / 6));
ctx.lineTo(arrowX + arrowLength * Math.cos(arrowDirection + Math.PI / 6), arrowY + arrowLength * Math.sin(arrowDirection + Math.PI / 6));
ctx.closePath();
ctx.fill();