问题1.找到与x轴正方向成以下角度的直线的斜率：
(i)-π/ 4
解决方案：
We have
θ = -π/4
Let the slope of the line be ‘m’
We know slope m = tan θ
m = tan (-π/4)
= -1
Therefore, Slope = -1
(ii)2π/ 3
解决方案：
We have
θ = 2π/3
Let the slope of the line be ‘m’
We know slope m = tan θ
m = tan (2π/3)
= tan (π – π/3)
= -tan π/3
= √3
Therefore, Slope = √3
(iii)3π/ 4
解决方案：
We have
θ = 3π/4
Let the slope of the line be ‘m’
We know slope m = tan θ
m = tan (3π/4)
= tan (π – π/4)
= -tan π/4
= -1
Therefore, Slope = -1
(iv)π/ 3
解决方案：
We have
θ = π/3
Let the slope of the line be ‘m’
We know slope m = tan θ
m = tan (π/3)
= tan π/3
= √3
Therefore, Slope = √3
问题2.找到通过以下点的直线的斜率：
(i)(-3、2)和(1、4)
解决方案：
Let ‘m’ be the slope of the line
So, the slope of line m = [4 – 2] / [1 – (-3)] [Using: m = [y2 – y1] / [x2 – x1]
= 2/4
= 1/2
(ⅱ)(在2 1，2原子1)和(以2 2，2原子2)
解决方案：
Slope of line ‘m’ = [2at2 – 2at1] / [at22 – at21]
= [2at2 – t1] / [at22 – t21]
= 2a(t2 – t1)/a(t2 – t1)(t2 + t1)
= 2/(t2 + t1)
(iii)(3，-5)和(1，2)
解决方案：
Slope of line m = [2 – (-5)] / [1 – 3] [Using: m = [y2 – y1] / [x2 – x1]
= 7/-2
= -7/2
问题3.说明以下各条中的两条线是平行，垂直还是不平行。
(i)通过(5，6)和(2，3)；到(9，-2)和(6，-5)
解决方案：
Let A(5, 6), B(2, 3), C(9, -2) and D(6, -5) be the given points. Then
m1 = Slope of the line AB
= [3 – 6] / [2 – 5]
= -3/-3
= 1
m2 = Slope of the line CD
= [-5 – (-2)] / [6 – 9]
= -3/3
= -1
Clearly, m1 m2 = -1
This shows that AB is perpendicular to CD
(ii)通过(9，5)和(-1，1);到(3，-5)和(8，-3)
解决方案：
Let A(9, 5), B(-1, 1), C(3, -5) and D(8, -3) be the given points. Then
m1 = Slope of the line AB
= [1 – 5] / [-1 – 9]
= -4/-10
= 2/5
m2 = Slope of the line CD
= [-3 – (-5)] / [8 – 3]
= 2/5
Clearly, m1 = m2
This shows that AB is parallel to CD
(iii)通过(6，3)和(1，1)；到(-2，5)和(2，-5)
解决方案：
Let A(6, 3), B(1, 1), C(-2, 5) and D(2, -5) be the given points. Then
m1 = Slope of the line AB
= [1 – 3] / [1 – 6]
= -2/-5 = 2/5
m2 = Slope of the line CD
= [-5 – 5] / [2 – (-2)]
= 10/4 = -5/2
Clearly, m1 m2 = -1
This shows that AB is perpendicular to CD
(iv)通过(3、15)和(16、6)；到(-5，3)和(8，2)
解决方案：
Let A(3, 15), B(16, 6), C(-5, 3) and D(8, 2) be the given points. Then
m1 = Slope of the line AB
= [6 – 15] / [16 – 3]
= -9/13
m2 = Slope of the line CD
= [2 – 3] / [8 – (-5)]
= -1/13
Clearly we don’t see any relation between m1 and m2
So, neither parallel nor perpendicular
问题4：找到一条线的斜率
(i)将第一象限角一分为二
解决方案：
We have Line bisects the first quadrant
We know if the line bisects in the first quadrant, then
the angle must be between line and the positive direction of x-axis.
Since, angle = 90°/2 = 45°
Using the formula,
Slope of the line, m = tan θ
Slope of the line for a given angle is m = tan 45°
So, m = 1
∴ The slope of the line is 1.
(ii)沿逆时针方向与y轴正方向成30°角。
解决方案：
We have, The line make an angle 30° with the positive direction of y-axis.
We know, angle between line and positive side of axis is 90° + 30° = 120°
Using formula,
Slope of the line, m = tan θ
Slope of the line for the given angle is m = 120°
So, m =- √3
∴ The slope of the line is – √3.
问题5.使用斜率方法，证明以下几点是共线的
(i)A(4，8)，B(5，12)，C(9，28)
解决方案：
Using the formula,
Slope of the line = [y2 – y1] / [x2 – x1]
Slope of the line AB = [12 – 8] / [5 – 4]
= 4/1 = 4
Slope of the line BC = [28 – 12] / [9 – 5]
= 16/4 = 4
Slope of the line CA = [8 – 28] / [4 – 9]
= -20/-5 = 4
Clearly, AB = BC = CA
∴ The given points are collinear.
(ii)A(16，-18)，B(3，-6)，C(-10，6)
解决方案：
Using the formula,
Slope of the line = [y2 – y1] / [x2 – x1]
Slope of the line AB = [-6 – (-18)] / [3 – 16]
= 12/-13 = -12/13
Slope of the line BC = [6 – (-6)] / [-10 – 3]
= 12/-13 = -12/13
Slope of the line CA = [6 – (-18)] / [-10 – 16]
= 12/-13 = -12/13
Clearly, AB = BC = CA
∴ The given points are collinear.
问题6. y的值是多少，以便穿过(3，y)和(2，7)的线与穿过(-1，4)和(0，6)的线平行?
解决方案：
Using the formula,
Slope of the line = [y2 – y1] / [x2 – x1]
Slope of the line joining point (-1, 4) and (0, 6) is
m1 = [6 – 4] / [0 – (-1)]
= 2
Slope of the line joining point (3, y) and (2, 7) is
m2 = [7 – y] / [2 – 3]
= y – 7
As the two lines are parallel m1 = m2
⇒ 2 = y – 7
⇒ y = 9
问题7.如果一条线的斜率是
(i)零
解决方案：
If slope = tanθ = 0
⇒ tanθ = tan0
⇒ θ = 0
When the slope of a line is zero then the line is parallel to x-axis.
(ii)正面
解决方案：
When the slope is positive then tanθ is positive and θ is acute angle
thus the line makes an acute angle (0<θ<π/2) with the positive x-axis.
(iii)负面
解决方案：
When the slope is negative then tanθ is negative and θ is obtuse angle
thus the line makes an obtuse angle (θ>π/2) with the positive x-axis.
问题8。证明连接(2，-3)和(-5，1)的线与连接(7，-1)和(0，3)的线平行。
解决方案：
Using the formula,
Slope of the line = [y2 – y1] / [x2 – x1]
Slope of the line joining point (2, -3) and (-5, 1) is
m1 = [1 – (-3)] / [(-5) – 2]
= 4/(-7) = -(4/7)
Slope of the line joining point (7, -1) and (0, 3) is
m2 = [3 – (-1)] / [ 0 – 7]
= 4/(-7) = -(4/7)
Since m1 = m2, the two lines are parallel
问题9。证明连接(2，-5)和(-2，5)的线垂直于连接(6，3)和(1，1)的线。
解决方案：
Using the formula,
Slope of the line = [y2 – y1] / [x2 – x1]
Slope of the line joining point (2, -5) and (-2, 5) is
m1 = [5 – (-5)] / [-2 – 2]
= -5/2
Slope of the line joining point (6, 3) and (1, 1) is
m2 = [1 – 3] / [1 – 6]
= 2/5
m1 × m2 = [-5/2] × [2/5]
= -1
∴ The two given lines are perpendicular to each other.
问题10.不使用毕达哥拉斯定理，请证明点A(0，4)，点B(1、2)和点C(3，3)是直角三角形的顶点。
解决方案：
Using the formula,
Slope of the line = [y2 – y1] / [x2 – x1]
Slope of the line AB = [2 – 4] / [1 – 0]
= -2
Slope of the line BC = [3 – 2] / [3 – 1]
= 1/2
Slope of AB slope of BC = (-2) × (1/2) = -1
∴ Angle between AB and BC = π/2
∴ ABC are the vertices of a right-angled triangle.
问题11。证明点(-4，-1)，(-2，-4)，(4、0)和(2、3)是矩形的顶点。
解决方案：
Let A(-4, -1), B(-2, -4), C(4, 0) and D(2, 3) be the vertices of the rectangle
Slope of the line AB = [-4 + 1] / [-2 + 4]
m1 = -3/2
Slope of the line BC = [0 + 4] / [4 + 2]
m2 = 2/3
Slope of the line CD = [3 – 0] / [2 – 4]
m3 = -3/2
Slope of the line AD = [3 + 1] / [2 + 4]
m4 = 2/3
We see m1 = m3 & m2 = m4
∴ AB || CD & BC || AD
Also, m1 × m2 = -1, m2 × m3 = -1 & m3 × m4 = -1
∴AB⊥BC, BC⊥CD and CD⊥AD
Thus given set of points are the vertices of a rectangle.