11类RD Sharma解决方案–第23章直线-练习23.6 |套装2

问题11。找到通过点(-4、3)的直线的方程，使得方程在轴之间的部分可被点按比例5：3整除。

解决方案：

As we know that the equation of the line is 

x/a + y/b = 1    …..(1)

It is given that point P(-4, 3) divides the line joining A(a, 0) and B(0, b) in the ratio of 5:3 so,

[3a/8, 5b/8] = (-4, 3)

3a = -4

a = -32/3 

5b/8 = 3

b = 24/5 

Now put the value of a and b in eq(1), we get

3x/-32 + 5y/24 = 1

Hence, the equation of the line is 9x – 20y + 96 = 0

问题12。找到一条通过点(22，-6)的线的方程，使得在x轴上的截距比在y轴上的截距大5。

解决方案：

As we know that the equation of the line is 

x/a + y/b = 1    …..(1)

So, 

a = b + 5

x/b + 5 + y/b = 1

It is given that the line passes through the point(22, -6) 

So, 22/b + 5 – 6/b = 1

22b – 6b – 30 = 0

b² -11b + 30 = 0

b = 5 or 6

a = 10 or 11

When b = 5 and a = 10, then the equation of line is 

x/10 + y/5 =1

x + 2y – 10 = 0

When b = 6 and a = 11, then the equation of line is 

x/11 + y/6 = 1

6x + 11y = 66

Hence, the equations of the line are x + 2y – 10 = 0 and 6x + 11y = 66

问题13。找到直线的方程，该方程穿过P(1，-7)并分别与A和B处的轴相交，因此4AP – 3BP = 0。

解决方案：

As we know that the equation of straight line is

y – y₁ = m(x – x₁)    …..(1)

The line passes through point P(1, -7) and meets the axes at A(a, 0) and B(0, b)

So,

AP/BP = 3/4

By using the section formula, we get

lx₂ + mx₁/l+m, ly₂ + my₁/l + m

= 3(0) + 4(a)/3 + 4 = 1

= 4a/7 =1

a = 4/7

= 3(a) + 4(0)/3 + 4 = -7 

= 3b/7 = -7

b = -49/3

Put A(7/4, 0) and B(0, -49/3) in eq(1), we get

y – y₁ = y₂ – y₁  / x₂ – x₁ (x – x₁)

y – 0 = (-49/3 – 0)/(0 – 7/4)(x – 7/4)

y = 28/3(x – 7/4)

Hence, the equation of the line is 3y – 28x – 49 = 0

问题14.找到通过点(2，2)的线的方程，并截去总和为9的轴上的截距。

解决方案：

As we know that the equation of the line is 

x/a + y/b = 1    …..(1)

and a + b = 9

So 

x/a + y/9 – 1 = 1

It is given that the line passes through the point (2, 2)

2/a + 2/(a – a) = 1

18 – 2a + 2a = 9a – a²

a² – 9a – 18 = 0

a = 6, 3

b = 3, 6

When b = 3 and a = 6, then the equation of line is 

x/6 + y/3 = 1

2x – y – 6 = 0

When b = 6 and a = 3, then the equation of line is 

x/3 + y/b = 1

x + 2y – 6 = 0

Hence, the equations of the line are 2x – y – 6 = 0 and x + 2y – 6 = 0

问题15。求出通过点P(2，6)并分别在点A和B切割坐标轴的直线方程，以使AP / BP = 2/3。

解决方案：

As we know that the equation of straight line is

y – y₁ = m(x – x₁)    …..(1)

The line passes through point P(2, 6) and meets the axes at A(a, 0) and B(0, b)

So, AP/BP = 2/3

By using the section formula 

x = lx₂ + mx₁/l + m, ly₂ + my₁/l + m

l : m = 2 : 3

= 2(0) + 3(a)/2 + 3 = 2

= 3a =10

a = 10/3

= 2(b) + 3(0)/2 + 3 = 6

= 2b = 30

b = 15

So, the points are A(10/3, 0) and B(0, 15)

So, the equation of the line AB is 

y – y₁ = y₂ – y₁ / x₂ – x₁ (x – x₁)

y – 0 = 15/(10/3) (x – 10/3)

y = -45/10(x – 10/3)

2y = -9x + 90/3

Hence, the equation of the line is 9x + 2y = 30

问题16。找到直线的等式，每个直线都通过点(3，2)，并在X和Y轴上切割截距a和b，使得a – b = 2。

解决方案：

As we know that the equation of the line is 

x/a + y/b = 1    …..(1)

and a – b = 2

a = 2 + b

It is given that the line passes through the point (3, 2)

So, 3/b +2 + 2/b = 1

3b + 2b + 4 = b² + 2b

b² – 3b – 4 = 0

b = 4 or -1

a = 6 or 1

When b = 4 and a = 6, then the equation of line is 

x/6 + y/4 = 1

2x + 3y = 12

When b = -1 and a = 1, then the equation of line is 

x/1 + y/-1 = 1

x – y = 1

Hence, the equations of the line are 2x + 3y = 12 and x – y = 1

问题17：找到通过原点的直线方程，并将直线2x + 3y = 6的部分三等分，该部分在轴之间截取。

解决方案：

According to the question

The equation of the line is 2x + 3y = 6 

This line cuts coordinate axis at A(3, 0) and B(0, 2)

AP/PB = 1/2,

and

AQ/QB = 2/1

Hence, the coordinates of P = (1 × 0 + 3 × 2)/3, (1 × 2 + 0)/3 = 1/3, 2/3

The coordinates of Q = (2 × 0 + 3 × 1)/3, (4 + 0)/3 = 3/3, 4/3 

So, the equation of OQ is:

y – 0 = (4/3 – 0/ 1 – 0) × (x – 0)

3y = 4x

And the equation of OP is:

y – 0 = (2/3 – 0/1/3 – 0) × (x – 0)

x – 3y = 0

问题18。找到通过(2,1)并将两轴之间的直线部分3x-5y = 15平分的直线方程。

解决方案：

According to the question

The equation of the line is 3x – 5y = 15

x/5 – y/3 = 1

and this line cuts axis at (5, 0) and (-3, 0)

So, the  position of AB intercepted between the axis is 1 : 1

p = (5/2, -3/2)

It is given that the equation of line passing through the point(2, 1),

So,

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

y – 1 = -5(x – 2)

Hence, the equation of the line  is 5x + y = 11

问题19：找到通过原点的直线方程，然后将直线轴ax + +的部分平分+ c = 0，并在坐标轴之间截取。

解决方案：

According to the question 

The equation of line is ax + by + c = 0

ax + by = -c

x/(-c/a) + y/(-c/b) = 1

c = ((-c/a) + 0)/2, (0 – (c/b))/2 

c = (-c/2a, -c/2b)

It is given that the equation of line passing through the point (0, 0) and c(-c/2a, -c/2b)

So, (y + c/2b) = (-c/2b) / (-c/2a) (x + c/2a)

 = -y/a + x/b = 0

Hence, the equation of the straight line is ax – by = 0

问题20.由坐标轴和一条直线形成的三角形的面积为6个正方形单位，斜边的长度为5个单位。找到直线的方程式。

解决方案：

As we know that the equation of the line is 

x/a + y/b = 1    

and it meets the axes at A(a, 0) and B(0, b)

Let us considered OAB is a triangle and area of the triangle is 6

1/2 × OA ×OB = 6

1/2 × a × b = 6

b = 6/a

It is given that the hypotenuse of triangle OAB is 5

So, (a)² + (b)² = (5)²

(a)² + (6/a)² = (5)²

a = 4, 3

So, b = 3, 4

When a = 4 and b = 3 so the equation of the line is:

x/4 + y/3 = 1 

When a = 3 and b = 4 so the equation of the line is:

x/3 + y/4 = 1