问题1： A线穿过点A(1、2)，与x轴成60°角，并在点P处与x + y = 6线相交。找到AP。
解决方案：
Given: (x1, y1) = A (1, 2) and θ = 60°
By using the formula, the equation of the line is given by:
(x – x1) / cosθ = (y – y1) / sinθ = r
When we substitute the values, we get:
(x – 1) / cos60° = (y – 1) / sin60° = r
(x – 1) / (1/2) = (y – 1) / (√3/2) = r
Where, r represents the distance of any point on the line from A (1, 2).
The coordinate of point P on this line are
(1 + r/2, 2 + √3r/2)
and, P lies on the line x + y = 6
1 + r/2 + 2 + √3r/2 = 6
r = 6 / (1 + √3) = 3(√3 – 1)
Therefore, the value of AP = 3(√3 – 1)
问题2：如果通过点P(3，4)的直线与x轴成角度π / 6并在Q处与直线12x + 5y + 10 = 0相遇，则求出长度PQ。
解决方案：
Given: (x1, y1) = A (3, 4) and θ = π/6 = 30°
By using the formula, the equation of the line is given by:
(x – x1) / cosθ = (y – y1) / sinθ = r
When we substitute the values, we get:
(x – 3) / cos30° = (y – 4) / sin30° = r
(x – 3) / (√3/2) = (y – 4) / (1/2) = r
x – √3 y + 4√3 – 3 = 0
Let PQ = r
The coordinate of Q are:
(x – 3) / cos30° = (y – 4) / sin30° = r
x = 3 + √3r/2, y = 4 + r/2
Given that the Q lies on 12x + 5y + 10 = 0
So,
12(3 + √3r/2) + 5(4 + r/2) + 10 = 0
66 + (12√3 + 5)r / 2 = 0
r = – 132 / 12√3 + 5
PQ = |r| = 132 / 12√3 + 5
Therefore, the length of PQ = 132 / 12√3 + 5.
问题3：通过点A(2，1)与x轴成正角成ππ / 4的直线在点B处与另一条直线x + 2y + 1 = 0相交。找到长度AB。
解决方案：
Given: (x1, y1) = A (2, 1) and θ = π/4 = 45°
By using the formula, the equation of the line is given by:
(x – x1) / cosθ = (y – y1) / sinθ = r
When we substitute the values, we get:
(x – 2) / cos45° = (y – 1) / sin45° = r
(x – 2) / 1/√2 = (y – 1) / 1/√2 = r
x – y – 1 = 0
Let AB = r
The coordinate of B are:
(x – 2) / cos45° = (y – 1) / sin45° = r
x = 2 + r/√2, y = 1 + r/√2
Given that the B lies on x + 2y + 1 = 0
So,
(2 + r/√2) + 2(1 + r/√2) + 1 = 0
5 + (3r/√2)r = 0
r = 5√2/3
Therefore, the length of AB = 5√2/3
问题4：穿过A(4，– 1)的直线a平行于直线3x – 4y + 1 =0。找到直线上与A的距离为5个单位的两个点的坐标。
解决方案：
Given: (x1, y1) = A (4, -1)
Also give equation of Line: 3x – 4y + 1 = 0
4y = 3x + 1
y = 3x/4 + 1/4
Slope tan θ = 3/4
Thus,
Sin θ = 3/5
Cos θ = 4/5
By using the formula, the equation of the line is given by:
(x – x1) / cosθ = (y – y1) / sinθ
When we substitute the values, we get:
(x – 4) / (4/5) = (y + 1) / (3/5)
3x – 4y = 16
Let, AP = r
and r = ±5
The coordinate of P are:
(x – 4) / (4/5) = (y + 1) / (3/5) = r
x = 4r/5 + 4 and y = 3r/5 + 4
x = 4(±5)/5 + 4 and y = 3(±5)/5 + 4
x = ±4 + 4 and y = ±3 –1
x = 8, 0 and y = 2, – 4
Therefore, the coordinates are (8, 2) and (0, −4) which are at the distance of 5 unit from the point A 4, -1).
问题5：与X轴成角度θ倾斜的P(x1，y1)直线与Q +中的c + 0直线ax +相交。求出PQ的长度。
解决方案：
Given that the line passes through P(x1, y1) and makes an angle of θ with the x–axis.
The equation of the line is given by:
(x – x1) / cosθ = (y – y1) / sinθ
Let PQ = r
The coordinates of Q are:
Now, by using the formula, the equation of the line is given by:
(x – x1) / cosθ = (y – y1) / sinθ = r
x = x1 + rcosθ , y = y1 + rsinθ
Given that the Q lies on ax + by + c = 0
So,
a(x1 + rcosθ) + b(y1 + rsinθ) + c = 0
r = PQ = | (ax1 + by1 + c) / (acosθ + bsinθ ) |
Therefore, the value of PQ = | (ax1 + by1 + c) / (acosθ + bsinθ) |
问题6：沿着与x轴成45°角的线，找到点(2，3)与2x – 3y + 9 = 0线的距离。
解决方案：
Given: (x1, y1) = (2, 3) and θ = 45°
By using the formula, the equation of the line is given by:
(x – x1) / cosθ = (y – y1) / sinθ = r
When we substitute the values, we get:
The coordinate are:
(x – 2) / cos45° = (y – 3) / sin45° = r
x = 2 + r/√2, y = 3 + r/√2
Given that the point lies on 2x – 3y + 9 = 0
So,
2(2 + r/√2) – 3(3 + r/√2) + 9 = 0
2[(2√2 + r)/√2] – 3[(3√2 + r)/√2] + 9 = 0
4√2 + 2r – 9√2 – 3r + 9√2 = 0
r = 4√2
Therefore, the distance is = 4√2 units.
问题7：求出点(3，5)距线2x + 3y = 14的距离，该线平行于斜率为1/2的线。
解决方案：
Given: (x1, y1) = (3, 5)
Also give equation of Line: 2x + 3y = 14
and Slope tan θ = 1/2
Thus,
Sin θ = 1/√5
Cos θ = 2/√5
By using the formula, the equation of the line is given by:
(x – x1) / cosθ = (y – y1) / sinθ = r
When we substitute the values, we get the coordinate of P as:
(x – 3) / (2/√5) = (y – 5) / (1/√5) = r
x = 2r/√5 + 3 and y = r/√5 + 5
Given that the point lies on 2x + 3y = 14
Therefore,
2(2r/√5 + 3) + 3(r/√5 + 5) = 14
4r/√5 + 6 + 3r/√5 +15 = 14
7r/√5 + 21 = 14
7r/√5 = 14 – 21
7r/√5 = -7
r = ±√5
Therefore, the distance is √5 units.
问题8：求出点(2，5)与直线3x + y + 4 = 0的距离，该距离平行于斜率为3/4的线进行测量。
解决方案：
Given: (x1, y1) = (2, 5)
Also give equation of Line: 3x + y + 4 = 0
and Slope tan θ = 3/4
Thus,
Sin θ = 3/5
Cos θ = 4/5
By using the formula, the equation of the line is given by:
(x – x1) / cosθ = (y – y1) / sinθ = r
When we substitute the values, we get the coordinate of P as:
(x – 2) / (4/5) = (y – 5) / (3/5) = r
x = 4r/5 + 2 and y = 3r/5 + 5
Given that the point lies on 3x + y + 4 = 0
Therefore,
3(4r/5 + 2) + 3r/5 + 5 + 4 = 0
12r/5 + 6 + 3r/5 +9 = 0
15r/5 + 15 = 0
15r/5 = -15
r = ±5
Therefore, the distance is 5 units.