如何找到一条线的三等分点？

几何是数学的一个分支，它处理线、角、线段、点等，帮助我们确定两个不同空间之间的空间关系。有许多类型的几何可以研究。其中一些是欧几里得几何、拓扑、球面几何、双曲几何、微分几何和射影几何。让我们了解一下线段，

线段

几何图形有两个端点，但没有厚度。可以测量线段的长度，但不能测量线。端点可以命名为例如 AB 来确定不同线段。

线段的属性

• 它们不是空集，而是连接的。
• 线段被视为有序几何的一部分。
• 线段的长度根据其图形固定，因为它有端点。

截面公式

这是一个属于坐标几何的主题。它用于求线段除以内部和外部点的比率。它还用于物理学中寻找系统的质心。我们主要研究它下面的三个公式：

1. 内部分工
2. 对外事业部
3. 中点公式

如何找到一条线的三等分点？

回答：

The formula in which a line is divided into three parts in a certain ratio of 1:2 or 2:1 internally. Use section formula to solve any problem. Section formula is mathematically given by

$(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n})$

Where m and n are the two integers of ratio given as m:n.

For the trisection formula, use the section formula twice,

Step 1: Solve the points of the line segment by using the ratio m:n = 1:2.

Step 2: Solve the points of the line segment by using the ratio m:n = 2:1.

Let’s take a look at an example, if the points are given are (3, 2) and (3, 4), according to the trisection rule, the line segment with points (3, 2) and (3, 4) will be divided into the ratios of 1:2 and 2:1.

Now,

(x₁, y₁) = (3, 2)

(x₂, y₂) = (3, 4)

For the ratio 1:2

m:n = 1:2

= $(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n})$

= (1x₃ + 2x₃/1 + 2, 1x₄ + 2x₂/1 + 2)

= ((3 + 6)/3 , (4 + 4)/3)

= (3, 8/3)

Then, for ratio 2:1

= $(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n})$

= (2x₃ + 1x₃/2 + 1, 2x₄ + 1x₂/2 + 1)

= ((6 + 3)/3, (8 + 2)/3)

= (3, 10/3)

示例问题

问题 1. 求点 (4,-2) 和 (3, 1) 的三等分点。

解决方案：

According to the trisection rule, the line segment with points (4,-2) and (3,1) will be divided into the ratios of 1:2 and 2:1.

Now,

(x1,y1)=(4,-2)

(x2,y2)=(3,1)

For the ratio 1:2

m:n=1:2

=>(mx2+nx1/m+n , my2+ny1/m+n)

=>(1×3+2×4/1+2, 1×1+2x(-2)/1+2)

=>(3+8/3, 1-4/3)

=>11/3, -1)

Then, for the ratio 2:1

m:n=2:1

=>(mx2+nx1/m+n , my2+ny1/m+n)

=>(2×3+1×4/2+1, 2×1+1x(-2)/2+1)

=>(6+4/3, 2-2/3)

=>(10/3, 0)

问题 2. 求点 (5, -6) 和 (-7, 5) 的三等分点。

解决方案：

According to the trisection rule, the line segment with points (5,-6) and (-7,5) will be divided into the ratios of 1:2 and 2:1.

Now,

(x1,y1)=(5,-6)

(x2,y2)=(-7,5)

For the ratio 1:2

m:n=1:2

=>(mx2+nx1/m+n , my2+ny1/m+n)

=>(1x(-7)+2×5/1+2, 1x(5)+2x(-6)/1+2)

=>(-7+10/3, 5-12/3)

=>(1,-7/2)

For the ratio 2:1

m:n=2:1

=>(mx2+nx1/m+n , my2+ny1/m+n)

=>(2x(-7)+1×5/2+1, 2x(5)+1x(-6)/2+1)

=>(-14+5/3, 10-6/3)

=>(-3,4/3)

问题 3. 求点 (2, 5) 和 (1, -8) 的三等分。

解决方案：

According to the trisection rule, the line segment with points (2,5) and (1,-8) will be divided into the ratios of 1:2 and 2:1.

Now,

(x1,y1)=(2,5)

(x2,y2)=(1,-8)

For the ratio 1:2

m:n=1:2

=>(mx2+nx1/m+n , my2+ny1/m+n)

=>(1×1+2×2/1+2, 1x(-8)+2×5/1+2)

=>(1+4/3,-8+10/3)

=>(5/3, 2/3)

For the ratio 2:1

m:n=2:1

=>(mx2+nx1/m+n , my2+ny1/m+n)

=>(2×1+1×2/2+1, 2x(-8)+1×5/2+1)

=>(2+2/3, -16+5/3)

=>(4/3,-11/3)