什么是垂直线公式？

垂线定义为通过另一条线上的^一点与原线成90°角的线。垂直线的斜率是原线斜率的倒数的负数。有两个关于垂直线的基本属性，可以推导出与垂直线相关的公式。让我们来看看他们，

垂直线公式

有为垂直线定义的属性。下面是两个性质，它们是斜率和垂线方程的乘积。

• 垂直线的斜率与原线斜率的乘积始终为-1 。

证明：

Lets the original line makes an angle of θ with the X-axis. Then, the line perpendicular to the line will make an angle of θ + 90° or θ – 90° with the X-axis.

Then, the slope of the original line is equal to tanθ.

The slope of the perpendicular line is equal to tan(θ + 90^o) or tan(θ – 90^o). 

Thus, the slope of the perpendicular line is -cotθ.

The product of the slopes = tanθ × (-cotθ) = -1

Hence, the product of the slopes is always equal to -1.

2.如果原直线的方程为ax + by + c = 0 ，则其垂线方程为– bx + ay + d = 0 ，其中 d 为常数。

证明：

The equation of the original line is ax + by + c = 0

The slope of the original line is -a / b.

Let’s the slope of the perpendicular line is m

Since product of the slopes is -1, so, we can write,

m × (-a / b) = – 1

m = b / a

So, if the perpendicular line passes through a point (x₁ , y₁), 

(y – y₁) / (x – x₁) = b / a

y – y₁ = (b / a) × (x – x₁)

ay – ay₁ = bx – bx₁

– bx + ay + (bx₁ – ay₁) = 0

Let’s bx₁ – ay₁ = d, where d is a constant.

Thus, the equation stands as,

– bx + ay + d = 0

示例问题

问题 1：线 3x + 2y + 5 = 0 和 2x – 3y + 8 = 0 是否垂直？

解决方案：

The slope of the line 3x + 2y + 5 = 0 is – 3 / 2.

The slope of the line 2x – 3y + 8 = 0 is -2 / (-3) = 2 / 3

Thus, the product of the slopes are: (- 3 / 2) × (2 / 3) = -1

Since, the product of slopes are -1, the lines are perpendicular.

问题 2：求垂直于线 x + 2y + 5 = 0 并通过点 (2, 5) 的线。

解决方案：

From property 2, we get that the equation of a line perpendicular to the line ax + by + c = 0 is – bx + ay + d = 0.

Comparing the line x + 2y + 5 = 0 with ax + by + c = 0, 

• a = 1
• b = 2
• c = 5

Thus, the equation of any line perpendicular to this line is – 2x + y + d = 0, where d is a constant.

Given, this line passes through (2, 5), thus putting (2, 5) in this equation of the perpendicular line, 

-2 × 2 + 5 + d = 0

• d = -1

Hence, the equation of the perpendicular line stands as -2x + y – 1 = 0

问题 3：求垂直于直线 3x + 9y + 7 = 0 的直线的斜率。

解决方案：

Given, the equation of the line is  3x + 9y + 7 = 0.

So, the slope of this line is – 3 / 9 = – 1 / 3.

Let’s, the slope of the perpendicular line be m.

From property 1, we can write,

m × (- 1 / 3) = – 1

m = 3

Thus, the slope of the line perpendicular to the given line is 3.

问题 4：求一条垂直于线 x + y + 3 = 0 的线与 X 轴在 [0, 90°] 范围内的角度。

解决方案：

Slope of the given line = – 1 / 1 = – 1

Lets, the slope of its perpendicular line be m. 

So, from property 1, we can write,

m × -1 = – 1

m = 1

So, if the angle of the line perpendicular to the given line is θ, then we can write slope as 

tanθ = 1

θ = tan^-1(1) = 45°

Hence, the angle made by perpendicular line with X-axis is 45°.