点梯度公式
笛卡尔平面中的一条直线经过无数个点。这些点中的每一个都有自己的 x 和 y 坐标。一条线通过的点用于找到它的斜率。不仅如此,这些点还可以用来写直线方程。下面讨论一种这样的方法。
点梯度公式
在许多以笛卡尔形式编写/查找/表达直线方程的方法中,点斜率或点梯度公式在坐标几何中占有非常重要的地位。顾名思义,这种形式的方程由直线经过的一个点及其斜率组成。
公式
点梯度公式如下:
y – y1 = m(x – x1)
Where,
- x and y depict general point coordinates.
- x1 and y1 are the numerical coordinates of a point through which the line passes.
- m represents the slope of the given line.
点梯度公式的推导
Slope of a line passing through two points (x, y) and (x1, y1) = m =
Multiplying both sides by (x – x1), we have:
⇒ y – y1 = m(x – x1)
Hence proved.
示例问题
问题 1:通过 (2, -4) 且斜率为 5 的直线的方程是什么?
解决方案:
The given point is (2, −4). Thus, x1 = 2, y1 = −4.
Also, m = slope = 5.
We know the point slope equation of a line is given by y – y1 = m(x – x1).
Substituting the above values in the equation, we have:
y – (-4) = 5(x – 2)
⇒ y + 4 = 5x − 10
⇒ y = 5x − 10 − 4
⇒ y = 5x − 14
问题 2:通过 (5, 2) 且斜率为 3/4 的直线的方程是什么?
解决方案:
The given point is (5, 2). Thus, x1 = 5, y1 = 2.
Also, m = slope = 3/4.
We know the point slope equation of a line is given by y – y1 = m(x – x1).
Substituting the above values in the equation, we have:
y − (2) = 3/4(x − 5)
⇒ y − 2 = 3x/4 − 15/4
⇒ y = 3x/4 − 15/4 + 2
⇒ y = 3x/4 − 7/4
问题 3:通过 (3, 3) 的水平线的方程是什么?
解决方案:
The given point is (3, 3). Thus, x1 = 3, y1 = −3.
Since the slope of a horizontal line is zero, m = 0.
We know the point slope equation of a line is given by y – y1 = m(x – x1).
Substituting the above values in the equation, we have:
y − (3) = 0(x − 3)
⇒ y – 3 = 0
⇒ y = 3
问题4:假设一条直线经过点(1, 1) 和(-2, 4)。使用点斜率公式找到它的方程。
解决方案:
We know the point slope equation of a line is given by y – y1 = m(x – x1).
In order to use the point- slope form, we need to calculate the slope of the line first.
Slope = m =
⇒ m = -1
Substituting the above values in the equation, we have:
y – (1) = -1(x – 1)
⇒ y − 1 = -x + 1
⇒ y = -x + 2
问题 5:通过 (0, 3) 且斜率为 8 的直线的方程是什么?
解决方案:
The given point is (0, 3). Thus, x1 = 0, y1 = 3.
Also, m = slope = 8.
We know the point slope equation of a line is given by y – y1 = m(x – x1).
Substituting the above values in the equation, we have:
y − 3 = 8(x – 0)
⇒ y − 3 = 8x
⇒ y = 8x + 3