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Class 11 RD Sharma solutions – Chapter 23 The Straight Lines- Exercise 23.6 | Set 2
• Last Updated : 30 Apr, 2021

### Question 11. Find the equation to the straight line which passes through the point (-4, 3) and is such that the portion of it between the axes is divisible by the point in the ratio 5:3.

Solution:

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

### Question 12. Find the equation of a line which passes through the point(22, -6) and is such that the intercept on x-axis exceeds the intercept on y-axis by 5.

Solution:

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

### Question 13. Find the equation of the line, which passes through P (1, -7) and meets the axes at A and B respectively so that 4AP – 3BP = 0.

Solution:

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

### Question 14. Find the equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.

Solution:

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

### Question 15. Find the equation of the straight line which passes through the point P (2, 6) and cuts the coordinate axes at the point A and B respectively so that AP/BP = 2/3.

Solution:

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

### Question 16. Find the equation of the straight lines each of which passes through the point (3, 2) and cuts the intercepts a and b on X and Y-axis such that a – b = 2.

Solution:

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

### Question 17. Find the equations of the straight lines which pass through the origin, and trisect the portion of the straight line 2x + 3y = 6, which is intercepted between the axes.

Solution:

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

### Question 18. Find the equation of the straight line passing through (2,1) and bisecting the portion of the straight line 3x-5y = 15 lying between the axes.

Solution:

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

### Question 19. Find the equation of the straight line passing through the origin and, bisecting the portion of the line ax + by + c = 0, intercepted between the coordinate axes.

Solution:

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

### Question 20. The area of the triangle formed by coordinate axes and a line is 6 square units and the length of the hypotenuse is 5 units. Find the equation of the line.

Solution:

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

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