问题1.将以下方程式简化为斜率-截距形式，并找到其斜率和y-截距。
(i)x + 7y = 0
(ii)6x + 3y – 5 = 0
(iii)y = 0
解决方案：
(i) x + 7y = 0
Given that,
The equation is x + 7y = 0
Slope – intercept form is represented as y = mx + c, where m is the slope and c is the y intercept
Therefore, the above equation can be represented as,
y = -1/7x + 0
Hence, the above equation is in the form of y = mx + c, where m = -1/7 and c = 0.
(ii) 6x + 3y – 5 = 0
Given that,
The equation is 6x + 3y – 5 = 0
Slope – intercept form is represented as y = mx + c, where m is the slope and c is the y intercept
Therefore, the above equation can be represented as,
3y = -6x + 5
y = -6/3x + 5/3 = -2x + 5/3
Hence, the above equation is in the form of y = mx + c, where m = -2 and c = 5/3.
(iii) y = 0
Given that,
The equation is y = 0
Slope – intercept form is represented as y = mx + c, where m is the slope and c is the y intercept
Therefore, the above equation can be represented as,
y = 0 × x + 0
Hence, the above equation is in the form of y = mx + c, where m = 0 and c = 0.
问题2.将以下方程式简化为截距形式，并在轴上找到其截距。
(i)3x + 2y – 12 = 0
(ii)4x – 3y = 6
(iii)3y + 2 = 0
解决方案：
(i) 3x + 2y – 12 = 0
Given that,
The equation is 3x + 2y – 12 = 0
As we know that equation of line in intercept form is x/a + y/b = 1, where a and b are intercepts on x axis and y axis respectively.
Therefore, 3x + 2y = 12
Let’s divide both sides by 12, and we will get
3x/12 + 2y/12 = 12/12
x/4 + y/6 = 1
Hence, the above equation is of the form x/a + y/b = 1, where a = 4, b = 6
Intercept on x axis = 4
Intercept on y axis = 6
(ii) 4x – 3y = 6
Given that,
The equation is 4x – 3y = 6
As we know that equation of line in intercept form is x/a + y/b = 1, where a and b are intercepts on x axis and y axis respectively.
Therefore, 4x – 3y = 6
let’s divide both sides by 6, and we will get
4x/6 – 3y/6 = 6/6
2x/3 – y/2 = 1
x/(3/2) + y/(-2) = 1
Hence, the above equation is of the form x/a + y/b = 1, where a = 3/2, b = -2
Intercept on x axis = 3/2
Intercept on y axis = -2
(iii) 3y + 2 = 0
Given that,
The equation is 3y + 2 = 0
As we know that equation of line in intercept form is x/a + y/b = 1, where a and b are intercepts on x axis and y axis respectively.
Therefore, 3y = -2
Let’s divide both sides by -2, and we will get
3y/-2 = -2/-2
3y/-2 = 1
y/(-2/3) = 1
Hence, the above equation is of the form x/a + y/b = 1, where a = 0, b = -2/3
Intercept on x axis = 0
Intercept on y axis = -2/3
问题3.将以下等式简化为正常形式。找到它们与原点的垂直距离以及垂直轴与正x轴之间的夹角。
(i)x –√3y+ 8 = 0
(ii)y – 2 = 0
(iii)x – y = 4
解决方案：
(i) x – √3y + 8 = 0
Given that,
The equation is x – √3y + 8 = 0
As we know that equation of line in normal form is given by x cos θ + y sin θ = p where θ is the angle between perpendicular and positive x axis and p is perpendicular distance from origin respectively.
Therefore, x – √3y + 8 = 0
x – √3y = -8
Let’s divide both sides by √(12 + (√3)2) = √(1 + 3) = √4 = 2, and we will get,
x/2 – √3y/2 = -8/2
(-1/2)x + √3/2y = 4
Now the above equation is in the form of x cos 120o + y sin 120o = 4
Hence, the above equation is of the form x cos θ + y sin θ = p, where θ = 120° and p = 4.
Perpendicular distance of line from origin = 4
Angle between perpendicular and positive x – axis = 120°
(ii) y – 2 = 0
Given that,
The equation is y – 2 = 0
As we know that equation of line in normal form is given by x cos θ + y sin θ = p where θ is the angle between perpendicular and positive x-axis and p is perpendicular distance from origin respectively.
Therefore, 0 × x + 1 × y = 2
Let’s divide both sides by √(02 + 12) = √1 = 1, and we will get,
0 (x) + 1 (y) = 2
Now the above equation is in the form of x cos 90o + y sin 90o = 2
Hence, the above equation is of the form x cos θ + y sin θ = p, where θ = 90° and p = 2.
Perpendicular distance of line from origin is 2 and
Angle between perpendicular and positive x – axis is 90°
(iii) x – y = 4
Given that,
The equation is x – y + 4 = 0
As we know that equation of line in normal form is given by x cos θ + y sin θ = p where θ is the angle between perpendicular and positive x-axis and p is perpendicular distance from origin respectively.
Therefore, x – y = 4
Let’s divide both sides by √(12 + 12) = √(1+1) = √2, and we will get,
x/√2 – y/√2 = 4/√2
(1/√2)x + (-1/√2)y = 2√2
Now the above equation is in the form of x cos 315o + y sin 315o = 2√2
Hence, the above equation is of the form x cos θ + y sin θ = p, where θ = 315° and p = 2√2.
Perpendicular distance of line from origin is 2√2 and
Angle between perpendicular and positive x – axis is 315°
问题4.找到点(–1，1)与直线12(x + 6)= 5(y – 2)的距离。
解决方案：
Given that,
The equation of the line is 12(x + 6) = 5(y – 2).
12x + 72 = 5y – 10
12x – 5y + 82 = 0 ——–(i)
Now, after comparing equation (i) with general equation of line Ax + By + C = 0, we get A = 12, B = –5, and C = 82
Perpendicular distance d of a line Ax + By + C = 0 from a point (x1, y1) is given by,
d = |Ax1 + By1 + C| / √A2 + B2
Given points (x1, y1) are (-1, 1)
Distance of point (-1, 1) from the given point is
d = |12 x (-1) + (-5) + 82| / √122 + (-5)2 = 65 / 13 units
= 5 units.
Hence, the distance is 5 units.
问题5.找到x轴上与线x / 3 + y / 4 = 1的距离为4个单位的点。
解决方案：
Given that,
The equation of line is x/3 + y/4 = 1
4x + 3y = 12
4x + 3y – 12 = 0 ——(i)
Now, after comparing equation (i) with general equation of line Ax + By + C = 0, we get A = 4, B = 3, and C = -12
Let’s assume that (a, 0) be the point on the x-axis, whose distance from the given line is 4 units.
Therefore, the perpendicular distance d of a line Ax + By + C = 0 from a point (x1, y1) is given by,
d = |Ax1 + By1 + C| / √A2 + B2
4 = |4a + 3 × 0 – 12| / √42 + 32
4 = |4a – 12| / 5
|4a – 12| = 4 × 5
± (4a – 12) = 20
4a – 12 = 20 or – (4a – 12) = 20
4a = 20 + 12 or 4a = -20 + 12
a = 32/4 or a = -8/4
a = 8 or a = -2
Hence, the required points on the x – axis are (-2, 0) and (8, 0)
问题6.找出平行线之间的距离
(i)15x + 8y – 34 = 0和15x + 8y + 31 = 0
(ii)l(x + y)+ p = 0和l(x + y)– r = 0
解决方案：
(i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0
Given that,
The parallel lines are 15x + 8y – 34 = 0 & 15x + 8y + 31 = 0.
By using the formula, the distance d between parallel lines Ax + By + C1 = 0 and Ax + By + C2 = 0 is given by,
d = |C1 – C2| / √A2 + B2
From given equation we get, A = 15, B = 8, C1 = -34, C2 = 31
Now apply the formula and calculate distance between parallel lines,
d = |-34 – 31| / √152 + 82 = |-65| / √225 + 64
= 65 / 17
Hence, the distance between parallel lines is 65/17.
(ii) l(x + y) + p = 0 and l (x + y) – r = 0
Given that,
The parallel lines are l (x + y) + p = 0 and l (x + y) – r = 0.
lx + ly + p = 0 and lx + ly – r = 0
By using the formula,
By using the formula, the distance d between parallel lines Ax + By + C1 = 0 and Ax + By + C2 = 0 is given by,
d = |C1 – C2| / √A2 + B2
From given equation we get, A = l, B = l, C1 = p, C2 = -r
Now apply the formula and calculate distance between parallel lines,
d = |p – (-r)| / √l2 + l2 = |p+ r| / √2 l = |p+r|/l√2
Hence, the distance between parallel lines is |p+r|/l√2
问题7.找到与线3x − 4y + 2 = 0平行并经过点(–2，3)的线的方程。
解决方案：
Given that,
The line is 3x – 4y + 2 = 0
Therefore, y = 3x/4 + 2/4 = 3x/4 + ½
The above equation is in the form of y = mx + c, where m is the slope of the given line.
Therefore, slope of the given line is 3/4
As we know that parallel line have same slope.
Therefore, slope of other line = m = 3/4
Equation of line having slope m and passing through (x1, y1) is given by, y – y1 = m (x – x1)
Now put the value of slope 3/4 and points (-2, 3) in above formula, and we get,
y – 3 = ¾ (x – (-2))
4y – 3 × 4 = 3x + 3 × 2
3x – 4y = 18
Hence, the equation is 3x – 4y = 18
问题8.求出与x – 7y + 5 = 0垂直且x截距为3的直线的方程。
解决方案：
Given that,
The equation of line is x – 7y + 5 = 0
Therefore, y = 1/7x + 5/7
The above equation is in the form of y = mx + c, where m is the slope of the given line.
Therefore, slope of the given line is 1/7
Slope of the line perpendicular to the line having slope m is -1/m,
Therefore, slope of the line perpendicular to the line having a slope of 1/7 is -1/(1/7) = -7.
The equation of line with slope -7 and x intercept 3 is given by y = m(x – d)
y = -7 (x – 3)
y = -7x + 21
7x + y = 21
Hence, the equation is 7x + y = 21
问题9.找出线√3x+ y = 1和x +√3y= 1之间的角度。
解决方案：
Given that,
The lines are √3x + y = 1 and x + √3y = 1
Therefore, y = -√3x + 1 ———(i) &
y = -1/√3x + 1/√3 ——–(ii)
Slope of line (i) is -√3, while the slope of line (ii) is -1/√3
Let’s assume that θ be the angle between two lines,
As we know that,
tanθ = |m1 – m2| / |1 + m1m2|
Put the values of m1 and m2 formula, and we will get,
= |-√3 – (-1/√3)| / |1 + (-√3)(-1/√3)| = 1 / √3
θ = 30°
Hence, the angle between the given lines is either 30° or 180°- 30° = 150°
问题10.通过点(h，3)和(4，1)的线与线7x − 9y -19 = 0相交。成直角。找出h的值。
解决方案：
Let’s assume that the slope of the line passing through (h, 3) and (4, 1) be m1,
Therefore, m1 = (1-3)/(4-h) = -2/(4-h)
Let’s slope of line 7x – 9y – 19 = 0 be m2
7x – 9y – 19 = 0
Therefore, y = 7/9x – 19/9
m2 = 7/9
Given that, the given lines are perpendicular
m1 × m2 = -1
-2/(4-h) × 7/9 = -1
-14/(36-9h) = -1
-14 = -1 × (36 – 9h)
36 – 9h = 14
9h = 36 – 14
h = 22/9
Hence, the value of h is 22/9.