在练习1至8中，找到满足给定条件的直线方程：
问题1.编写x轴和y轴的方程式。
解决方案：
The y-coordinate of every point on x-axis is 0 and the x-coordinate of every point on y-axis is 0
So, the equation of x-axis is y = 0 and the equation of y-axis is y = 0.
问题2.以1/2的坡度穿过点(– 4，3)。
解决方案：
Given that point p(x1, y1) is (-4, 3) and slope m = 1/2
Equation of Line can be derived by formula y – y1 = m (x – x1),
Where m is slope of line and (x1, y1) is co-ordinate of p from which line passes.
y – 3 = 1/2(x – (-4))
y – 3 = 1/2 (x + 4)
2(y – 3) = x + 4
2y – 6 = x + 4
x + 4 – (2y – 6) = 0
x + 4 – 2y + 6 = 0
x – 2y + 10 = 0
So, the equation of the line is x – 2y + 10 = 0.
问题3.以(m，0)穿过(0，0)。
解决方案：
Given that Point p(x1, y1) is (0, 0) and slope is m.
So, the equation of Line can be derived by formula y – y1 = m (x – x1),
Where m is slope of line and (x1, y1) is co-ordinate from which line passes.
So, y – 0 = m (x – 0)
y = mx
y – mx = 0
So, the equation of the line is y – mx = 0.
问题4到(2，2√3)传递和在75的角度与所述x轴倾斜。
解决方案：
Given that point p(x1, y1) is (2, 2√3) and θ = 75°
So, the equation of line is (y – y1) = m (x – x1)
where, m = slope of line = tan θ and (x1, y1) are the points through which line passes
So, m = tan 75°
Now, finding tan 75° using below formula:
tan(A + B) = (tanA + tanB)/(1 – tanA.tanB)
tan 75° = tan(45° + 30°)
tan 75° = (tan 45° + tan 30°)/(1 – tan 30°.tan 45°)
tan 75° = (1 + 1/√3)/(1 – 1/√3)
tan 75° = (√3 + 1)/(√3 – 1)
By rationalizing, we get
tan 75° = 2 + √3
Equation of Line will be,
(y – 2√3) = (2 + √3)(x – 2)
y – 2√3 = 2 x – 4 + √3x – 2 √3
y = 2 x – 4 + √3x
(2 + √3)x – y – 4 = 0
So, the equation of the line is (2 + √3)x – y – 4 = 0.
问题5.将x轴与原点的左侧相距3个单位，斜率为–2。
解决方案：
Given that Slope m = –2, x-intercept = 3,
that means Line is passing from point p(x1, y1) that is (3, 0)
Equation of line will be
y – 0 = (–2) × (x − 3).
y = (–2) × (x + 3)
y = –2x – 6
2x + y + 6 = 0
So, the equation of the line is 2x + y + 6 = 0.
问题6.与y轴相交于原点上方2个单位的距离，并与x轴的正方向成30°角。
解决方案：
Given that θ = 30° then slope(m) will be = tan θ
m = tan30° = (1/√3)
As Y intercept is 2 that means line is passing from (0, 2) point p(x1, y1) = (0, 2)
Equation of Line will be
y – 2 = (1/√3)x
y = (1/√3)x + 2
y = (x + 2√3) / √3
√3 y = x + 2√3
x – √3 y + 2√3 = 0
So, the equation of the line is x – √3 y + 2√3 = 0.
问题7.通过点(–1、1)和(2，–4)。
解决方案：
Given that point p1(x1, y1) is (–1, 1) and point p2(x2, y2) is (2, –4),
It is mention in question that line passes from p1 and p2.
That means, the Slope(m) of line will be (y2 – y1)/(x2 – x1)
m = (–4 – 1)/(2 – (–1))
m = –5/3
Equation of Line will be
(y – y1) = m(x – x1)
y – 1 = –5/3 (x + 1)
3 (y – 1) = (–5)(x + 1)
3y – 3 = –5x – 5
3y – 3 + 5x + 5 = 0
5x + 3y + 2 = 0
So, the equation of the line is 5x + 3y + 2 = 0.
问题8.与原点的垂直距离是5个单位，垂直线与x轴正方向形成的角度为30°。
解决方案：
Given that p = 5 and ω = 30°
We know that the equation of the line having normal distance p from the origin and
angle ω which the normal makes with the positive direction of x-axis is given by x cos ω + y sin ω = p.
On substituting the values in the equation, we get
x cos30° + y sin30° = 5
x(√3 / 2) + y(1/2) = 5
√3 x + y = 5(2) = 10
√3 x + y – 10 = 0
So, the equation of the line is √3 x + y – 10 = 0.
问题9.ΔPQR的顶点是P(2，1)，Q(–2，3)和R(4，5)。通过顶点R找到中位数的方程式。
解决方案：
Given Vertices of ΔPQR that are P (2, 1), Q (–2, 3) and R (4, 5)
Let RS be the median of vertex R => S is a midpoint of PQ.
As S is midpoint of PQ => S = (P + Q)/2
S = (2 – 2, 1 + 3)/2
S = (0, 2)
Equation of the line passing through the points (x1, y1) and (x2, y2) is given by
y – 5 = –3/ –4(x – 4)
(–4)(y – 5) = (–3)(x – 4)
–4y + 20 = –3x + 12
–4y + 20 + 3x – 12 = 0
3x – 4y + 8 = 0
So, the equation of median through the vertex R is 3x – 4y + 8 = 0.
问题10。找到通过(–3，5)并垂直于通过点(2，5)和(–3，6)的直线的方程。
解决方案：
Given that Points are (2, 5) and (-3, 6).
So, the slope, m1 = (y2 – y1)/(x2 – x1)
= (6 – 5)/(–3 – 2)
= 1/–5 = –1/5
As we know that two non-vertical lines are perpendicular to each other
if their slopes are negative reciprocals of each other.
Then, m = (–1/m1)
= –1/(–1/5)
= 5
As we know that the point p(x, y) lies on the line with
slope m through the fixed point (x1, y1),
If its coordinates satisfy the equation y – y1 = m (x – x1)
Then, y – 5 = 5(x – (–3))
y – 5 = 5x + 15
5x + 15 – y + 5 = 0
5x – y + 20 = 0
So, the equation of the line is 5x – y + 20 = 0