第11类RD Sharma解决方案–第23章直线-练习23.19

问题1.通过线4x – 3y = 0和2x – 5y + 3 = 0且与4x + 5 y + 6 = 0平行的线的交点求出一条直线方程。

解决方案：

From the question we have,

The equations of lines which are :

 4x – 3y = 0 ………….. (1)

2x – 5y + 3 = 0 …….. (2)

Now we find the equation of the straight line that pass through the

points of intersection of equations (1) and (2) is given,

4x − 3y + λ (2x − 5y + 3) = 0 ……………………… (3)

(4 + 2λ)x + (− 3 − 5λ)y + 3λ = 0……………….. (4)

y = (4 + 2λ / 3 + 5λ)x + 3λ/(3 + 5λ)

The line is parallel to 4x + 5y + 6 = 0 or, y = -4x/5 – 6/5

4 + 2λ / 3 + 5λ = -4/5

λ = -16/15

Now put the value of λ in eq(4), we get

(4 – (32/15))x – (3 – (80/15))y – 48/15 = 0

Hence, the equation of line is 28x + 35y – 48 = 0

问题2。找到一条直线方程，该方程通过x + 2y + 3 = 0和3x + 4y + 7 = 0的交点并垂直于直线x – y + 9 = 0。

解决方案：

From the question we have,

The equations of lines which are :

x + 2y + 3 = 0 ….. (1)

3x + 4y + 7 = 0  …. (2)

Now we find the equation of the straight line that pass through the

points of intersection of equations (1) and (2) is given,

So, x + 2y + 3 + λ(3x + 4y + 7) = 0 ………….. (3)

(1 + 3λ)x + (2 + 4λ)y + 3 + 7λ = 0 …………. (4)

y = – ((1 + 3λ)/(2 – 4λ))x – ((3 + 7λ)/(2 + 4λ))

So, the slope of the line is -(1 + 3λ/2 + 4λ)

It is given that the line whose slope is -(1 + 3λ/2 + 4λ)

is perpendicular to the straight line x – y + 9 = 0

So, -(1 + 3λ/2 + 4λ) × 1 = 1

λ = 1

Now put the value of λ in eq(3), we get

x + 2y + 3 + 1(3x + 4y + 7) = 0

x + y + 2 = 0

Hence, the equation of the line is x + y + 2 = 0

问题3。找到通过2x – 7y + 11 = 0和x + 3y – 8 = 0的交点并与(i) x –轴(ii)y –轴平行的线的方程。

解决方案：

From the question we have,

The equations of lines which are :

 2x – 7y + 11 = 0 ….. (1)

x + 3y – 8 = 0 ….. (2)

Now we find the equation of the straight line that pass through the

points of intersection of equations (1) and (2) is given,

2x − 7y + 11 + λ(x + 3y − 8) = 0 ………. (3)

(2 + λ)x + (− 7 + 3λ)y + 11 − 8λ = 0 ………. (4)

(i) x – axis

2 + λ = 0

λ = -2

Now, we put the value of λ in equation (4), we get

0 + (− 7 − 6)y + 11 + 16 = 0

13y − 27 = 0

Hence, the equation of the required line is 13y − 27 = 0

(2) y – axis

7 + 3λ = 0

λ = 7/3

Now, we put the value of λ in equation (4), we get

(2 + 7/3)x + 0 + 11 – 8(7/3) = 0

13x – 23 = 0

Hence, the equation of the required line is 13x – 23 = 0

问题4.找到穿过2x + 3y + 1 = 0和3x – 5y – 5 = 0且相对于轴倾斜的交点的直线方程。

解决方案：

From the question we have,

The equations of lines which are :

 2x + 3y + 1 = 0  ….. (1)

 3x – 5y – 5 = 0 ….. (2) 

Now we find the equation of the straight line that pass through the

points of intersection of equations (1) and (2) is given,

2x + 3y + 1 + λ(3x − 5y − 5) = 0   …………….. (3)

(2 + 3λ)x + (3 − 5λ)y + 1 − 5λ = 0  …………… (4)

y = – [(2 + 3λ) / (3 – 5λ)] – [(1 – 5λ) / (3 – 5λ)]

It is given that the line is equally inclined to the axes. 

So, the slope of the line is either 1 or − 1.

Hence,

– [(2 + 3λ) / (3 – 5λ)] = 1 and – [(2 + 3λ) / (3 – 5λ)] = -1

-2 – 3λ = 3 – 5λ and 2 + 3λ = 3 – 5λ

λ = 5/2 and 1/8

Now, we put the value of λ in equation (4), we get

= (2 + 15/2)x + (3 – 25/2)y + 1 – 25/2 = 0 and,

= (2 + 3/8)x + (3 – 5/8)y + 1 – 5/8 = 0

19x – 19y – 23 = 0 and 19x + 19y + 3 = 0

Hence, the  equation of line is 19x – 19y – 23 = 0 and 19x + 19y + 3 = 0

问题5。找到通过直线x + y = 4和2x – 3y = 1的交点绘制的直线方程，该直线垂直于截断轴上截距5、6的直线。

解决方案：

From the question we have,

The equations of lines which are :

 x + y = 4  …..(1)

2x – 3y = 1 …..(2) 

Now we find the equation of the straight line that pass through the

points of intersection of equations (1) and (2) is given,

x + y − 4 + λ(2x − 3y − 1) = 0   …(3)

(1 + 2λ)x + (1 − 3λ)y − 4 − λ = 0     ….(4)

y = – [(1 + 2λ) / (1 – 3λ)]x + [(4 + λ) / (1 – 3λ)]

It is given that the equation of the line with intercepts 5 and 6 on the axis is

x/5 + y/6 = 1 ……..(5)

and the slope is -6/5

The lines (4) and (5) are perpendicular so,

-6/5 × [(-1 + 2λ) / (1 – 3λ)] = -1

λ = 11/3

Now, put the values of λ in (1), we get 

(1 + 2(11/3))x + (1 – 3(11/3))y − 4 – 11/3 = 0

(1 + 22/3)x + (1 – 11)y – 4 – 11/3 = 0

25x – 30y – 23 = 0

Hence, the equation of line is 25x – 30y – 23 = 0

问题6.证明由x(1 +λ)+ y(2 –λ)+ 5 = 0表示的线族通过固定点，λ是任意的。另外，找到固定点。

解决方案：

According to the question

The family of lines represented by x (1 + λ) + y (2 – λ) + 5 = 0

So, x + xλ + 2y – λy + 5 = 0

λ(x – y) + (x + 2y + 5) = 0

(x + 2y + 5) + λ(x – y) = 0

So, this is L₁ + λL₂ = 0

Hence, the line passing through the intersection of x – y = 0 and x + 2y = -5.

So, (-5/3, -5/3) which is the fixed point through which the lines pass for any value of λ.

问题7：证明对于不同的k值，由(2 + k)x +(1 + k)y = 5 + 7k给出的直线经过一个固定点。另外，找到要点。

解决方案：

From the question we have,

The equation of line is :

(2 + k) x + (1 + k) y = 5 + 7k

(2x + y – 5) + (x + y = 7) = 0

It is of the form l₁ + kL₂ = 0

So, it represents a line passing through:

2x + y – 5 = 0 ……(1)

x +y – 7 = 0 ……(2)

On Solving equation(1) and (2) we get, 

The value of the point (-2, 9). 

问题8.找到通过2x + y – 1 = 0和x + 3y – 2 = 0的交点并用坐标轴使面积为3/8 sq单位的三角形的直线方程。

解决方案：

From the question we have,

The equations of lines which are :

2x + y – 1 = 0    …..(1)

x + 3y – 2 = 0     …..(2)

Now we find the equation of the straight line that pass through the

points of intersection of equations (1) and (2) is given,

(2x +y -1) + λ(x + 3y – 2) = 0     ……(3)

x (2 + λ) + y (1 + 3λ) – 1 – 2λ = 0

x/((1 + 2λ) / (2 + λ)) + 4/((1 + 2λ) / (1 + 3λ)) = 1

As we know that the area of the triangle OAB = 1/2 × OB × OA

8 / 3 =1/2 × (y intercept) × (x intercept)

8/3 = 1/2 × (1 + 2λ / 1 + 3λ) × (1 + 2λ / 2 + λ)

16/3 = (4λ² + 4λ) / (3λ + 3λ² + 7λ)

60λ² + 124λ + 35 = 0

λ = -124 ± √(124)² – 4×60 × 35 / 2×60

λ = -124 ± √15376-8400 / 120

λ = 1(Approximately)

Now put the value of λ in (3) we get 

3x + 4y -30 = 0,

Hence, the equation of line is  12x + y – 3 = 0

问题9.找到通过直线3x – y = 5和x + 3y = 1的交点的直线方程，并在轴上做相等且正的截距。

解决方案：

From the question we have,

The equations of lines which are :

3x – y = 5   …..(1)

x + 3y = 1     …..(2)

Now we find the equation of the straight line that pass through the

points of intersection of equations (1) and (2) is given,

(3x – y – 5) + λ(x + 3y – 1) = 0    …..(3)

x/(5 + λ/3 + λ) + y/(5 + λ/3λ – 1) = 1

It is given that the line which makes equal and positive intercepts with the line

So, 5 + λ /3 + λ = 5 + λ/3λ – 1

3λ – 1 = 3 + λ

2λ = 4

λ = 2

Now put the value of λ in (3) we get 

3x – y – 5 + 2x + 6y – 2 = 0

Hence, the equation of line is 5x + 5y = 7

问题10.求出直线方程通过线x – 3y + 1 = 0和2x – 5y – 9的交点，其与原点的距离为√5。

解决方案：

From the question we have,

The equations of lines which are :

 x – 3y + 1 = 0  …..(1)

2x – 5y – 9      …..(2)

Now we find the equation of the straight line that pass through the

points of intersection of equations (1) and (2) is given,

x – 3y + 1 + λ(2x + 5y – 9) = 0    ….(3)

(1 + 2λ) x + (-3 + 5λ) y + 1 – 9λ = 0

It is given that the distance from the origin is√5

So, 

D = |((1 + 2λ) 0 + (-3 + 5λ) 0 + 1 – 9λ) / √(1 + 2λ)² + (5λ – 3)² |

√5  = | 1 – 9λ / √1 + 4λ² + 4λ + 25λ² + 9 – 30λ |

5(10 + 29λ² – 26λ) = (1 – 9λ)²

50 + 145λ² – 130λ = 1 + 81λ² – 18λ²

64λ² – 112λ + 49 = 0

(8λ – 7)² = 0 

λ =7/8

Now put the value of λ in (3) we get 

= x – 3y + 1+ (7/8) (2x + 5y – 9) = 0

8x – 24y + 8 + 14x + 35y – 63 = 0

22x + 11y – 55 = 0

Hence, the equation of line is 2x + y – 5 = 0

问题11。通过直线x – y + 1 = 0和直线2x – 3y + 5 = 0的交点找到直线方程，该直线与点(3，2)的距离为7/5。

解决方案：

From the question we have,

The equations of lines which are :

x – y + 1 = 0 …..(1)

2x – 3y + 5 = 0      …..(2)

On solving these two equations of lines we get,

intersection point (2, 3).

Now, let equation of a line passing through (2, 3) be 

y – mx + c

3 = 2m + c

c = 3 – 2m

So, the equation of the line is y – mx + 3 – 2m       ……(3)

It is given that the distance from the point (3, 2) is 7/5.

|3m – 2 + 3 – 2m / √m² + 1 | = 7/5

| m + 1 / √m² +1 | = 7/5

(m + 1)² / m²+ 1 = 49 / 25

25(m² + 2m + 1) = 49m² + 49

25m² + 50m + 25 = 49m² + 49

24m² – 50m + 24 = 0

12m² – 25m + 12 = 0

m = 4/3, m = 3 /4

Now put the value of m in eq(3), we get

y = (4/3)x + 3 – 8/3

3y = 4x + 1

4x – 3y + 1 = 0

 y = (3/ 4)x +3 – 6/4

4y – 3y + 1 = 0

Hence, the equations of lines are 4x – 3y + 1 = 0 and 4y – 3y + 1 = 0