第11类RD Sharma解决方案-第23章直线-练习23.3

问题1.找到一条与x轴成150度角并从y轴截取截距2的直线的方程。

解决方案：

Given, slope, m = tan (150^°) ⇒ m = -1/√3, also, y-intercept = (0,2)

We know that equation, of a line is given as y = mx+ c, m is the slope and c is the intercept that line cuts on y-axis, therefore

equation for the line will be:

y =  -x/√3 + 2

⇒ x – 2√3 + √3y = 0

问题2.找出直线方程：

(i)具有斜率2和y截距3；

(ii)坡度为-1 / 3，y轴截距为-4

(iii)坡度为-2，且x轴与原点左侧相距3个单位。

解决方案：

(i) We know that equation, of a line is given as y = mx+ c, therefore equation for the line with slope 2 and y-intercept 3 will be: y = 2x + 3

(ii) Similarly, equation of the line with slope -1/3 and y-intercept -4 will be: y = -x/3 – 4

⇒ x +3y +12 = 0

(iii) Since, the line cuts the x-axis at a distance 3 units to the left of origin its coordinate will be (-3,0) and the given slope, m = -2.

Equation of a line passing through a point is given by the formula: y-y₁  = m (x – x₁), hence the equation will be

⇒ y – 0 = -2(x – (-3))

⇒ y = -2x -6

⇒ 2x + y + 6 = 0

问题3.找到坐标轴之间的角度的平分线方程。

解决方案：

The equation of the line on the coordinate axes are x=0 and y=0.

The equations of the bisectors of the angle between  x=0 and y=0 are:

x ± y = 0

问题4。找到一条直线方程，该方程^{与x轴成tan -1} (3)的角度，并在y轴的负方向上切掉4个单位的截距。

解决方案：

Here, ∅ = tan^-1 (3) ⇒ m = tan∅ =3, line cuts cuts off an intercept of 4 units on negative direction of y-axis, so the coordinate will be (0, -4)

Hence, the equation of the line is: y = 3x -4

问题5。找到一条y截距为-4且与连接(2，-5)和(1,2)的线平行的方程。

解决方案：

Since, our required line is parallel to the line passing through the coordinates (2, -5) and (1,2), it will have the same slope as the later line. Therefore, slope, m = (y₂ –  y₁) / (x₂ – x₁) = (2 – (-5) ) / (1 – 2 ) = -7

Also, given y-intercept = -4, hence the required equation of the line is: y = -7x – 4

问题6.找到一条直线方程，该方程垂直于连接(4,2)和(3,5)的线，并在y轴上切掉长度为3的截距。

解决方案：

Slope of the line passing through the points (4,2) and (3,5) is

(y2 –  y1) / (x2 – x1) = (5 – 2 ) / (3 – 4 ) = -3

Now, our required line is perpendicular to the former line, so its slope will be m = 1/3

Also, y-intercept, c = 3, hence the required equation of the line is: y = x/3 + 3

⇒ x – 3y +9 = 0

问题7.如果它与y轴相交截距-3，则求出与(4,3)和(-1,1)线垂直的方程。

解决方案：

Slope of the line passing through the points (4,3) and (-1,1) is

(y2 –  y1) / (x2 – x1) = (1 – 3 ) / (-1 – 4 ) = 2/5

Now, our required line is perpendicular to the former line, so its slope will be m = -5/2

Also, y-intercept, c = -3, hence the required equation of the line is: y = -5x/2 – 3

⇒ 5x + 2y +6 = 0

问题8.在距原点以上2个单位的距离上找到与y轴相交的直线的方程，并与x轴的正方向成30°角。

解决方案：

Given, m = tan 30° = 1/√3

Since, the line intersects the y-axis at a distance 2 units above the origin its coordinate will be (0,2)

Equation of a line passing through a point is given by the formula: y-y1  = m (x – x₁), hence the equation will be

⇒ y – 2 = 1/√3 . (x – 0)

⇒ √3y – 2√3 = x

⇒ x – √3y + 2√3 = 0