问题1：找到一条线的方程
(i)p = 5，α= 60°
(ii)p = 4，α= 150°
(iii)p = 8，α= 225°
(iv)p = 8，α= 300°
解决方案：
(i) p = 5, α = 60°
Given: p = 5, α = 60°
By using the formula of the equation of line, x cos α + y sin α = p, and substituting the values, we get
x cos 60° + y sin 60° = 5
x/2 + √3y/2 = 5
x + √3y = 10
Therefore, the equation of line is x + √3y = 10.
(ii) p = 4, α = 150°
Given: p = 4, α = 150°
By using the formula of the equation of line, x cos α + y sin α = p, and substituting the values, we get
x cos 150° + y sin 150° = 4
x cos(180° – 30°) + y sin(180° – 30°) = 4 [As, cos (180° – θ) = – cos θ , sin (180° – θ) = sin θ]
– x cos 30° + y sin 30° = 4
–√3x/2 + y/2 = 4
-√3x + y = 8
Therefore, the equation of line is -√3x + y = 8.
(iii) p = 8, α = 225°
Given: p = 8, α = 225°
By using the formula of the equation of line, x cos α + y sin α = p, and substituting the values, we get
x cos 225° + y sin 225° = 8
– x/√2 – y/√2 = 8
x + y + 8√2 = 0
Therefore, the equation of line is x + y + 8√2 = 0
(iv) p = 8, α = 300°
Given: p = 8, α = 300°
By using the formula of the equation of line, x cos α + y sin α = p, and substituting the values, we get
x cos 300° + y sin 300° = 8
x/2 – y√3/2 = 8
x – √3y = 16
Therefore, the equation of line is x – √3y = 16
问题2：找到直线的方程，在该方程上，从原点到直线的垂直线段的长度为4，垂直线段与x轴正方向的倾斜度为30°。
解决方案：
Given: p = 4, α = 30°
By using the formula of the equation of line, x cos α + y sin α = p, and substituting the values, we get
x cos 30° + y sin 30° = 4
x√3/2 + y1/2 = 4
√3x + y = 8
Therefore, the equation of line is √3x + y = 8.
问题3：找到与原点的垂直距离为4个单位的法线，法线与x轴正方向的夹角为15°。
解决方案：
Given: p = 4, α = 15°
By using the formula of the equation of line, x cos α + y sin α = p, and substituting the values, we get
x cos 15° + y sin 15° = 4
Now as, cos 15° = cos (45° – 30°) = cos45°cos30° + sin45°sin30° [Since, cos (A – B) = cos A cos B + sin A sin B ]
= 1/√2 × √3/2 + 1/√2 × 1/2
= 1/2√2( √3 + 1 )
And sin 15 = sin (45° – 30°) = sin 45° cos 30° – cos 45° sin 30° [Since, Sin (A – B) = sin A cos B – cos A sin B ]
= 1/√2 × √3/2 – 1/√2 × 1/2
= 1/2√2(√3 – 1)
By using the formula of the equation of line, x cos α + y sin α = p, and substituting the values, we get
x × [1/2√2(√3 + 1)] + y × [1/2√2(√3 – 1)]
(√3+1)x +(√3-1) y = 8√2
Therefore, the equation of line is (√3+1)x +(√3-1) y = 8√2.
问题4：找到距原点3个单位的直线方程，以使原点到直线的垂线与x轴正方向的夹角为tanα= 5/12。
解决方案：
Given: p = 3, α = tan-1 (5/12)
tan α = 5/12
So,
sin α = 5/13
cos α = 12/13
By using the formula of the equation of line, x cos α + y sin α = p, and substituting the values, we get
12x/13 + 5y/13 = 3
12x + 5y = 39
Therefore, the equation of line is 12x + 5y = 39.
问题5：找到直线的方程，在直线上，距原点的垂线的长度为2，且垂线与x轴的夹角为α，使得sinα = 1/3。
解决方案：
Given: p = 2, sin α = 1/3
As, cos α = √(1 – sin2 α)
= √(1 – 1/9)
= 2√2/3
By using the formula of the equation of line, x cos α + y sin α = p, and substituting the values, we get
x2√2/3 + y/3 = 2
2√2x + y = 6
Therefore, the equation of line is 2√2x + y = 6.
问题6：找到直线的方程，在直线上，距原点的垂线的长度为2，而垂线的斜率为5/12。
解决方案：
Given: p = ±2
tan α = 5/12
Therefore, sin α = 5/13
cos α = 12/13
By using the formula of the equation of line, x cos α + y sin α = p, and substituting the values, we get
12x/13 + 5y/13 = ±2
12x + 5y = ±26
Therefore, the equation of line is 12x + 5y = ±26.
问题7：从原点到直线的垂线长度为7，直线与y轴正方向成150°角。找到直线的方程式。
解决方案：
Given: p = 7 (perpendicular distance from origin)
Also given that the angle made with y-axis is 150°
therefore, the angle made with x-axis is 180° – 150° = 30°
sin 30° = 5/13
cos 30° = 12/13
By using the formula of the equation of line, x cos α + y sin α = p, and substituting the values, we get
x(√3/2) + y(1/2) = 7
√3x + y = 14
Therefore, the equation of line is √3x + y = 14
问题8：如果方程xcosθ+ysinθ = p id是直线的法线√3x+ y + 2 = 0，则找到θ和p的值
解决方案：
Given equation of line = √3x + y + 2 = 0
Which can also be written as -√3x – y = 2
(-√3/2)x + (-1/2)y = 1
This is same as the equation of line i.e. x cos α + y sin α = p
Therefore, cosθ = -√3/2
sinθ = -1/2
p = 1
Hence, θ = 210° = 7π/6 and p =1
问题9：求出直线方程，该方程使面积为96√3的三角形与轴成直角，并且与原点垂直的直线与y轴成30°角。
解决方案：
Given: Perpendicular from origin makes an angle of 30° with y-axis.
Therefore, it makes 60° with the x-axis.
Also given area of triangle = 96√3
1/2 × 2p × 2p/√3 = 96√3
p2 = (96√3 × √3) / 2 = 48 × 3 = 144
p = 12
By using the formula of the equation of line, x cos α + y sin α = p, and substituting the values, we get
x cos60° + y sin60° = 12
(1/2)x + (√3/2) = 12
x + √3y = 24
Therefore, the equation of line is x + √3y = 24