第 11 课 RD Sharma 解决方案-第 23 章直线-练习 23.2

问题 1. 求平行于 x 轴并通过 (3, -5) 的直线的方程。

解决方案：

Let the equation of line be: 

y – y1 = m(x – x1)

As line is parallel to x-axis, so slope(m) of the line would be equal to 0 (i.e. m=0).

Here (x1, y1) = (3, -5)

So, 

y – (-5) = 0 (x – 3)

y + 5 = 0

So, the required equation of the line is y + 5 = 0 or y = -5.

问题 2. 求与 x 轴垂直且在 x 轴上截距为 -2 的直线的方程。

解决方案：

Any line perpendicular to x-axis will have slope = -1/0

Let the equation of line be: 

y – y1 = m(x – x1) 

Here m = -1/0 and (x1, y1)= (-2, 0)

y – 0 = (-1/0)(x – (-2))

y – 0 = (-1/0)(x + 2)

-(x + 2) = 0

x + 2 = 0

So the required equation of the line is x + 2 = 0 or x = -2.

问题 3. 求与 x 轴平行且在 y 轴上截距为 -2 的直线的方程。

解决方案：

The slope of x-axis is 0 and line parallel to x-axis will also have same slope, therefore m = 0

Line has intercept -2 on y-axis

Therefore, (x1, y1) = (0, -2)

Let the equation of line be :

y – y1 = m(x – x1)

y – (-2) = 0(x – 0)

y + 2 = 0

So, the required equation of the line is y + 2 = 0 or y = -2.

问题 4. 画线 x = -3, x = 2, y = -2, y = 3，写出这样形成的正方形的顶点坐标。

解决方案：

The figure with the lines x = -3, x = 2, y = -2, y = 3 is as follows: 

From the above figure, we can say that the coordinates of the vertices of the square are:

(2, 3), (-3, 3), (-3, -2), (2, -2).

问题 5. 求通过 (4, 3) 且分别平行和垂直于 x 轴的直线方程。

解决方案：

Slope of line parallel to x-axis = 0

As the line passes through (4, -3)

The required equation of the line parallel to x-axis is 

y – y1 = m(x – x1) 

y – (3) = 0(x – 4)

y – 3 = 0

y = 3

Slope of a line perpendicular to x-axis = -1/0

The required equation of the line perpendicular to x-axis is 

y – y1 = m(x – x1)

y – 3 = (-1/0)(x – 4)

x – 4 = 0

x = 4

问题 6. 求与线 x = -2 和 x = 6 等距的线的方程。

解决方案：

The lines x = -2 and x = 6 pass through the points (-2, 0) and (6, 0) respectively.

Let (h, k) be the mid-point of the line joining the points (-2, 0) and (6, 0).

Therefore, (h, k)=((-2 + 6) / 2, 0) = (2, 0)

The given lines are parallel to the y-axis and the required line is equidistant from

these lines. Hence, the required line is parallel to the y-axis, which is given by x = k.

This line passes through (2, 0)

Therefore, 2 – k = 0

k = 2

So, the equation of a line that is equidistant from the lines x = -2 and x = 6 is x = 2. 

问题 7. 求直线 y = 10 和 y = -2 等距直线的方程。

解决方案：

The lines y = 10 and y = -2 pass through the points (0, 10) and (0, -2) respectively.

Let (h, k) be the mid-point of the line joining the points (0, 10) and (0, -2).

Therefore, (h, k) = (0, (10 – 2) / 2) = (0, 4)

The given lines are parallel to the x-axis and the required line is equidistant from

these lines. Hence, the required line is parallel to the y-axis, which is given by y = k.

This line passes through (0, 4)

Therefore, k = 4

So, the equation of a line that is equidistant from the lines y = 10 and y = -2 is y = 4.