第 11 课 RD Sharma 解决方案 - 第 23 章直线 - 练习 23.9
问题 1：将方程 √3x + y + 2 = 0 简化为：
(i) 斜率截距形式并找到斜率和 y 截距
(ii) 截距形式并在轴上找到截距
(iii) 范式并找到 p 和 α。
解决方案：
(i) slope-intercept form and find slope and y-intercept
Given equation:√3x + y + 2 = 0
Therefore, y = – √3x – 2
Thus, m = -√3, c = -2
Hence, the slope = – √3 and y-intercept = -2
(ii) Intercept form and find intercept on the axes
Given equation: √3x + y + 2 = 0
√3x + y = -2 [ Divide both sides by -2 ]
√3x/-2 + y/-2 = 1
Therefore, x-intercept = -2/√3 and y-intercept = -2
(iii) The normal form and find p and α
Given equation: √3x + y + 2 = 0
-√3x – y = 2 [ Divide both sides by 2 ]
(-√3/2)x – y/2 = 1
The normal form is represented as x cos α + y sin α = p
Thus,
cos α = -√3/2 = cos 210°
sin α = -1/2 = sin 210°
Therefore, p = 1 and α = 210°
问题 2：将以下方程化简为范式，并在每种情况下找到 p 和 α：
(i) x + √3y – 4 = 0
(ii) x + y + √2 = 0
(iii) x – y + 2√2 = 0
(iv) x – 3 = 0
(v) y – 2 = 0
解决方案：
(i) x + √3y – 4 = 0
Given equation: x + √3y – 4 = 0
x + √3y = 4 [ Divide both sides by 2 ]
(1/2)x + (√3/2)y = 2
The normal form is represented as x cos α + y sin α = p
Thus,
cos α = 1/2 = cos 60°
sin α = √3/2 = sin 60°
Therefore, p = 2 and α = 60°
(ii) x + y + √2 = 0
Given equation: x + y + √2 = 0
-x – y = √2 [ Divide both sides by √2 ]
(-1/√2)x – (1/√2)y = 1
The normal form is represented as x cos α + y sin α = p
Thus,
cos α = -1/√2
sin α = -1/√2
Since, both are negative,
Thus α is in III quadrant,
α = π(π/4) = 5π/4 = 225°
Therefore, p = 1 and α = 225°
(iii) x – y + 2√2 = 0
Given equation: x – y + 2√2 = 0
-x + y = 2√2 [ Divide both sides by √2 ]
(-1/√2)x – (-1/√2)y = 2
The normal form is represented as x cos α + y sin α = p
Thus,
cos α = -1/√2
sin α = -1/√2
α is in II quadrant,
α = (π/4) + (π/2) = 3π/4 = 135°
Therefore, p = 2 and α = 135°
(iv) x – 3 = 0
Given equation: x – 3 = 0
x = 3
The normal form is represented as x cos α + y sin α = p
Thus,
cos α = 1 = cos 0°
Therefore, p = 3 and α = 0°
(v) y – 2 = 0
Given equation: y – 2 = 0
y = 2
The normal form is represented as x cos α + y sin α = p
Thus,
sin α = 1 = sin π/2°
Therefore, p = 2 and α = π/2°
问题 3：将方程 x/a + y/b = 1 化为斜率截距形式，求其斜率和 y 截距？
解决方案：
Given equation: x/a + y/b = 1
Since, the general equation of line is represented as y = mx + c.
Thus,
bx + ay = ab
ay = – bx + ab
y = -bx/a + b
Hence, m = -b/a, c = b
Therefore, the slope = -b/a and y-intercept = b
问题 4：将线 3x – 4y + 4 = 0 和 2x + 4y – 5 = 0 简化为范式，从而找出哪条线更靠近原点？
解决方案：
Given equations:
3x − 4y + 4 = 0 …… (i)
2x + 4y − 5 = 0 …… (ii)
For equation (i),
-3x + 4y = 4 [ Divide both sides by 5 i.e. √(-3)2 + (4)2 ]
(-3/5)x + (4/5)y = 4/5
Therefore, p = 4/5
Now for equation (ii),
2x + 4y = – 5
-2x – 4y = 5 [ Divide both sides by √20 i.e. √(-2)2 + (-4)2 ]
(-2/√20)x – (4/√20)y = 5/√20
Therefore, p = 5/√20 = 5/4.47
Comparing value of p for equations (i) and (ii) we get,
4/5 < 5/4.47
Therefore, the line 3x − 4y + 4 = 0 is nearest to the origin.
问题 5：证明原点与线 4x + 3y + 10 = 0 等距； 5x – 12y + 26 = 0 和 7x + 24y = 50？
解决方案：
Given equations:
4x + 3y + 10 = 0 …… (i)
5x – 12y + 26 = 0 …… (ii)
7x + 24y = 50 …… (iii)
For equation (i),
4x + 3y + 10 = 0
-4x – 3y = 10 [ Divide both sides by 5 i.e. √(-4)2 + (-3)2 ]
(-4/5)x – 3/5)y = 10/5
(-4/5)x – 3/5)y = 2
Therefore, p = 2
For equation (ii),
5x − 12y + 26 = 0
-5x + 12y = 26 [ Divide both sides by 13 i.e. √(-5)2 + (12)2 ]
(-5/13)x + (12/13)y = 26/13
(-5/13)x + (12/13)y = 2
Therefore, p = 2
For equation (iii),
7x + 24y = 50 [ Divide both sides by 25 i.e. √(7)2 + (24)2 ]
(7/25)x + (24/25)y = 50/25
(7/25)x + (24/25)y = 2
Therefore, p = 2
Hence, the origin is equidistant from the given lines.
问题 6：求 θ 和 p 的值，如果方程 x cos θ + y sin θ = p 是直线 √3x + y + 2 = 0 的范式？
解决方案：
Given equation: √3x + y + 2 = 0
√3x + y = 2
-√3x – y = 2
The normal form is represented as x cos θ + y sin θ = p
Thus,
cos θ = -√3
sin θ = 1
tan θ = 1/√3
θ = π+(π/6) = 180° + 30° = 210°
Therefore, p = 2 and θ = 120°
问题 7：将方程 3x – 2y + 6 = 0 化简为截距，求 x 截距和 y 截距？
解决方案：
Given equation: 3x – 2y + 6 = 0
-3x + 2y = 6 [ Divide both sides by 6 ]
(-3/6)x + (2/6)y = 6/6
(-1/2)x + (1/3)y = 1
Therefore, x-intercept = -2 and y-intercept = 3
问题8：直线到原点的垂直距离为5个单位，斜率为-1。求直线方程？
解决方案：
Given: Perpendicular distance from the origin to the line is 5 units i.e p = 5
The normal form is represented as x cos α + y sin α = p
x cos α + y sin α = 5
y sin α = -x cos α + 5
y = (-x (cos α/sin α) + 5)
y = -x cot α + 5
Comparing it with the equation y = mx + c,
m = -cot α
-1 = -cot α
cot α = 1
α = π/4
Thus, the equation is,
x cos π/4 + y sin π/4 = 5
x/√2 + y/√2 = 5
x + y = 5√2
Therefore, x + y = 5√2 is the equation of line.