第11类RD Sharma解决方案–第23章直线-练习23.15

问题1.找到点(4，5)与直线3x – 5y + 7 = 0的距离

解决方案：

From question, we have,

The line: 3x – 5y + 7 = 0

By comparing ax + by + c = 0 and 3x − 5y + 7 = 0, 

a = 3, b = − 5 and c = 7

So, the distance of the point (4, 5) from the straight line 3x − 5y + 7 = 0 is

d = |(3 × 4 – 5 × 5 + 7)/√3² + (-5²) |

Hence, the distance (d) is 6/√34

问题2。找到从原点开始连接点(cosθ，sinθ)和(cos ϕ，sin ϕ)的线的垂直距离。

解决方案：

According to the question

Points are: (cos θ, sin θ) and (cosϕ, sin ϕ).

The equation of the line joining the points (cos θ, sin θ) and (cos ϕ, sin ϕ),

y – sin θ = (sinϕ – sin θ) / (cosϕ – cos θ) (x – cos θ)

(cosϕ – cos θ)y – sin θ(cosϕ – cos θ) = (sinϕ – sin θ)x – (sinϕ – sin θ)cos θ

Let us considered D be the perpendicular distance from the origin to the line

(sinϕ – sin θ)x – (cosϕ – cos θ)y + sinθcosϕ – sinϕcos θ = 0

D = |( sin θ – ϕ)/ √ (sinϕ – sin θ)² + (cosϕ – cos θ)²)|

= |(sin θ – ϕ) / sin²ϕ + sin²θ – 2sinϕsin θ + cos²ϕ + cos²θ – 2cos ϕcos θ|

= |1/√2 (sin (θ – ϕ)) / √1 – (cos (θ – ϕ))|

On solving these we get,

D = cos (θ – ϕ/2)

Hence, the distance is cos (θ – ϕ/2)

问题3.求出从原点到直线的垂直线的长度，这两个点的坐标为(a cosα，sinα)和(a cosβ，sinβ)。

解决方案：

According to the question

The coordinates are (a cos α, a sin α) and (a cos β, a sin β).

So, the equation line passing through (a cos α, a sin α) and (a cos β, a sin β) is,

 y – a sin α = ((a sin β – a sin α) / (a cos β – a cos α)) (x – a cos α)

 y – a sin α = ((sin β – sin α)/(cos β – cos α))(x – a cos α)

 y – a sin α = – cot((α – β)/2)(x – a cos α)

xcot((α – β)/2) + y – a sin α – a cos α cot((α –  β)/2) = 0

So, the distance of the line from the origin is

D = |(- a sin α – a cos α cot((α – β)/2)) / √cot²((α – β)/2) +1| 

= a|sin((α – β)/2)sin α + cos αcos((α – β)/2)|

= a |cos(((α – β)/2) – α)|

D = acos((α- β)/2)

Hence, the distance is acos((α- β)/2)

问题4：证明两条直线24x + 7y = 20和4x – 3y – 2 = 0上，垂直直线从直线2x + 11y – 5 = 0的任意点掉落。

解决方案：

According to the question

The equations of lines are  

24x + 7y = 20 

4x – 3y – 2 = 0

Now let us considered that Q(a, b) be any point lies on 2x + 11y − 5 = 0

So,

2a + 11b − 5 = 0

b = (5 – 2a)/11 ….. (1)

Let D1 and D2 be the perpendicular distance from point P on line 24x + 7y = 20 and 4x – 3y – 2 = 0,

So, D1 =| (24a + 7b – 20)/ √24² + 7²|

D1 = |((24a + 7) × ((5 – 2a)/11) – 20)/25)|

From (1),

D1 = | (50a – 37)/55 |

Similarly,

D2 = | (4a – 3b – 2)/ √3² + (-4)²|

From (1) we get,

D2 = | (50a – 37)/55 |

D1 = D2

Hence proved

问题5.找到直线2x + 3y = 21和直线3x – 4y + 11 = 0与直线8x + 6y + 5 = 0的交点距离。

解决方案：

According to the question

The equations of lines are  

 2x + 3y = 21    ……….. (1)   

 3x – 4y + 11 = 0   ………… (2)

By solving the eq(1) and (2) we get :-

x/(33 – 84) = y/(-63 – 22) = 1/(-8 – 9)

x = 3, 

y = 5

Hence, the point is (3, 5).

Now, we find the perpendicular distance D of the line 8x + 6y + 5 = 0 from the point (3, 5) is:

d = |(24 + 30 + 5)/√8² + 6²| = 59/10 = 5.9

Hence, the distance is 5.9

问题6.找到从点(4，-7)到连接原点和直线2x – 3y + 14和5x + 4y – 7 = 0的交点的线的垂直线的长度。

解决方案：

According to the question

The equations of lines are

2x – 3y + 14 = 0 

5x + 4y – 7 = 0

On solving both the equation we get

x = -35/23

y = 252/69

So, the point is (-35/23, 252/69)

and the equation of line joining origin and the point is y = mx,

where m = (252/69) / (-35/23) = -12/5

So, the equation of required line is y = -12x/5

12x + 5y = 0

Now, the perpendicular distance from (4, -7) to 12x + 5y = 0 is

D = |(12(4) + 5(-7)) / √12² + (-5)² |

D = 13/13

D = 1

Hence, the distance is 1

问题7.在x轴上与直线x / a + y / b = 1的垂直距离是a的点是什么?

解决方案：

Let us considered a point on x-axis is (a, 0)

So, the perpendicular distance from a line bx + ay = ab is,

a = |(ax₁ + by₁ + c)√a² + b²| 

Where,

a = b, b = a, x = -ab, x₁ = ±a, y₁ = 0

a = |(b(x) + a(0) – ab)/√a² + b²|

a = 0

or

a = (b(x) + a(0) – ab)/√a² + b²

a = (b/a)x = ±√a² + b²

x = 0

问题8.显示该垂线的上线(X / a)的COSθ+(Y / B)罪θ产物-从点^1(±√A2 – B ^2，0)为b ^2。

解决方案：

According to the question

We have to prove that the perpendicular distance from (±√a² – b², 0) to (x/a)cos θ + (y/b)sin θ – 1 is b²

So, D1 is the perpendicular distance from (√a² – b², 0) to (x/a)cos θ + (y/b)sin θ – 1 is

D1 = | ((√a² – b²/a)cos θ + (0x/b)sin θ – 1) / (cos² θ/a²) + (sin² θ/b²) |

= ((√a² – b²/a)cos θ – 1) / (cos² θ/a²) + (sin² θ/b²)           ….(1)

Also, D2 is the perpendicular distance from (-√a² – b², 0) to (x/a)cos θ + (y/b)sin θ – 1 is 

D2 = |((-√a² – b²/a)cos θ + (0x/b)sin θ – 1) / (cos² θ/a²) + (sin² θ/b²) |       ……(2)

So, from (1) and (2) we get

((√a²-b²/a²)cos² θ-1) / (cos² θ/a²)+(sin² θ/b²) = b²

Hence proved

问题9.求直线x –√3y+ 4 = 0上点(1、2)的垂线与原点的垂直距离。

解决方案：

According to the question

It is given that the perpendicular of (1, 2) on the straight line x – √3y = -4 

So, the equation is

y – y₁ = m'(x – x₁)

1 x₁ = 1, y₁ = 2, m = 1/√3, m’ = -√3

y – 2 = -√3 (x – 1)

y + √3x – (2 + √3) = 0……     (1)

It is given that the perpendicular distance from (0, 0) to eq(1) is

D = |(ax₁ + by₁ + c)| / √a² + b²

 a = √3, b = 1, c = -(2 + √3)

x₁ = 0, y₁ = 0

D = |√3 (0) + 1 (0) + (−2 − √3) | / (√3)² + 1² 

Hence, the distance is (2 + √3)/2

问题10。找到点(1、2)与具有斜率5的直线并经过x + 2y = 5和x – 3y = 7的交点的距离。

解决方案：

According to the question

The equations of lines are

 x + 2y = 5

x – 3y = 7

On solving both the equations we get a point A(29/5, -2/5)

So, the equation of line passing through point A(29/5, -2/5) which has slop 5 is 

y + (2/5) = 5(x – 29/5)

5y + 2 = 25x – 145 

25x – 5y – 147 = 0

So, D is the distance of (1, 2) from 25x – 5y – 147 = 0 is

D = | (25(1) – 5(2) – 147) / √25² + 5² |

|-132 √650|

Hence, the distance is |132 √650|

问题11.在y轴上与线的距离为x / 3 + y / 4 = 1等于4个单位的点是什么?

解决方案：

Let us assume that the point on the y-axis is (0, a)

It is given that the distance of (0, a) from line 4x + 3y – 12 = 0 is 4 units.

So, by using distance formula, we get

d = |(ax₁ + by + c)/√a² + b² |

4 = |(4(0) + 3(a) – 12)/√4² + 32 |

4 =|(3а – 12)/5|

4 = (-3a + 12)/5

-3a = 20 – 12

a = -8/3

And 4 =(3a – 12)/5

3a = 20 + 12

a = 32/3

Hence, the points are (0, 32/3), (0, -8/3)

问题12.在具有顶点A(2，3)，B(4，-1)和C(1，2)的三角形ABC中。从顶点A找到方程式和海拔高度的长度。

解决方案：

According to the question

ABC is a triangle whose vertices are A(2, 3), B(4, -1), and C(1, 2)

So, the equation of BC is

y + 1 = ((2 + 1/()1 – 4))(x – 4) 

y + 1 = -x + 4

x + y – 3 = 0

and AL = |(2 + 3 – 3)/√1 + 1 |

AL = √2

Here we conclude that the slope of line BC -1. So, the slope of AL is 1.

And it passes through A(2, 3) so, its equation is

y – 3 = 1(x – 2)

x – y + 1 = 0

问题13。表明一个移动点的路径，使其与两条线3x – 2y = 5和3x + 2y = 5的距离相等，是一条直线。

解决方案：

According to the question

The equations of lines are

3x – 2y = 5

3x + 2y = 5

Let us assume point P(h, k) is the moving point and equidistant from both the lines

So, 

|(3h – 2k – 5)/√9 + 4| = |(3h + 2k – 5)/√9 + 4|

|3h – 2k – 5| = |3h + 2k – 5|

4k = 0 or 6h – 10 = 0

k = 0

3h = 5

So, the locus of (h, k) is :

y = 0

or

3x = 5, which are straight lines.

问题14.如果变量点P(x，y)与直线x + y – 5 = 0和3x – 2y = 0的垂直距离的总和始终为10，则表明P必须在直线上移动。

解决方案：

According to the question

The sum of the perpendicular distances of a variable point P(x, y) 

from the lines (x + y – 5) = 0 and 3x – 2y + 7 = 0 is always 10

So, 

((x + y – 5)/√2) + ((3x – 2y+7) +√13) = 10 

(3√2 + √13)x + (√13 – 2√2) + (7√2 – 5√13 – 10√26) = 0

So, from here we conclude that it is a straight line.

问题15.如果垂直线的长度形成轴ax的点(1，1)– + c等于1。证明1 / c + 1 / a – 1 / b = c / 2ab。

解决方案：

Given that, the length of perpendicular from point (1, 1) to ax – by + c is 1

So, we have to prove that 1/c + 1/a – 1/b = c/2ab

By using the distance formula, we get

|(a (1) – b (1) + c)/√a² + b² | = 1

a – b + c = √a² + b²

(a – b + c)² = a² + b²

a² + b² + c² + 2ac – 2bc – 2ab = a² + b²

c² + 2ac – 2bc = 2ab

c + 2a – 2b = 2ab/c

(c/2ab) + (2a/2ab) – (2b/2ab) = 1/c

(c/2ab) = 1/c + 1/a – 1/b

Hence proved