📚 A-Level Mathematics: Distance Between Two Points and Area Calculation | A-Level 数学:两点距离与面积计算
The coordinate plane is formed by a horizontal x-axis and a vertical y-axis intersecting at the origin. Any point is represented by an ordered pair (x, y). This system lets us describe geometric properties such as distance and area in algebraic terms.
坐标平面由水平的 x 轴和竖直的 y 轴在原点相交而成。任意一点用有序数对 (x, y) 表示。借助这个坐标系,我们可以用代数方法描述距离、面积等几何性质。
1. The Coordinate Plane and the Distance Formula | 坐标平面与两点距离公式
Given two points A(x₁, y₁) and B(x₂, y₂), the straight-line distance between them is found using the distance formula. This formula is derived directly from Pythagoras’ theorem and is one of the most important tools in coordinate geometry.
已知两点 A(x₁, y₁) 和 B(x₂, y₂),它们之间的直线距离可以通过距离公式求得。该公式直接由勾股定理推导而来,是解析几何中最重要的工具之一。
d = √[(x₂ – x₁)² + (y₂ – y₁)²]
Here d is the length of the line segment joining the two points. You may also see this written as AB = √[(x₂ – x₁)² + (y₂ – y₁)²].
其中 d 是连接两点的线段长度。这个公式也可以写成 AB = √[(x₂ – x₁)² + (y₂ – y₁)²]。
2. Deriving the Distance Formula | 距离公式的推导
Place A(x₁, y₁) and B(x₂, y₂) on the coordinate plane. Draw a horizontal line from A and a vertical line from B so that they meet at a third point C(x₂, y₁). The triangle ACB is right-angled at C.
在坐标平面中取出 A(x₁, y₁) 和 B(x₂, y₂)。从 A 作水平线,从 B 作竖直线,两条线相交于点 C(x₂, y₁)。这样构成的三角形 ACB 在点 C 处是直角三角形。
The horizontal distance AC is |x₂ – x₁|, and the vertical distance CB is |y₂ – y₁|. The required distance AB is the hypotenuse of this right-angled triangle.
水平距离 AC 等于 |x₂ – x₁|,竖直距离 CB 等于 |y₂ – y₁|。我们所求的距离 AB 就是这个直角三角形的斜边。
By Pythagoras’ theorem:
根据勾股定理:
d² = (x₂ – x₁)² + (y₂ – y₁)²
Taking the positive square root gives the distance formula. The use of squared values means that the order of subtraction does not matter.
两边取正平方根,就得到距离公式。由于使用了平方运算,相减的先后顺序并不影响结果。
3. Worked Example: Distance in Action | 示例:两点距离的计算
Example 1: Find the distance between A(2, 1) and B(5, 5).
例 1:求点 A(2, 1) 与 B(5, 5) 之间的距离。
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Horizontal difference: x₂ – x₁ = 5 – 2 = 3.
水平差:x₂ – x₁ = 5 – 2 = 3。
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Vertical difference: y₂ – y₁ = 5 – 1 = 4.
竖直差:y₂ – y₁ = 5 – 1 = 4。
d = √(3² + 4²) = √25 = 5
So the distance between A and B is 5 units.
所以 A、B 两点之间的距离为 5 个单位。
Example 2: Find the distance between C(-3, 2) and D(1, -1).
例 2:求点 C(-3, 2) 与 D(1, -1) 之间的距离。
d = √[(1 – (-3))² + (-1 – 2)²] = √[4² + (-3)²] = √25 = 5
Notice how care with negative signs is essential. The horizontal difference is 4 and the vertical difference is -3, but squaring removes the negative sign.
注意处理负号非常重要。水平差为 4,竖直差为 -3,但平方运算会消除负号。
4. The Area of a Triangle by Base and Height | 用底和高求三角形面积
If you know the length of a base and the corresponding perpendicular height of a triangle, the area is straightforward to calculate.
如果已知三角形的底边长度以及对应的高,那么面积计算非常直接。
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