# Class 11 RD Sharma solutions – Chapter 23 The Straight Lines- Exercise 23.6 | Set 1

### Question 1. Find the equation to the straight line :

### (1) Cutting off intercepts 3 and 2 from the axes

**Solution**

Given that, a = 3, b = 2

Now we find the equation of line cutoff intercepts from the axes,

According to the formula of the equation of the line

x/a + y/b = 1

we get x/3 + y/2 = 1

By taking LCM,

2x + 3y = 6

Hence, the equation of line cut off intercepts 3 and 2 from the axes is 2x + 3y = 6

### (2) Cutting off intercepts -5 and 6 from the axes

**Solution**

Given that,

a = -5, b = 6

According to the formula of the equation of the line

x/a + y/b = 1

We get x/-5 + y/6 = 1 ….(1)

Take LCM of eq(1)

6x â€“ 5y = -30

Hence, the equation of line cut off intercepts 3 and 2 from the axes is 6x â€“ 5y = -30

### Question 2. Find the equation of the straight line which passes through (1, -2) and cuts off equal intercepts on the axes.

**Solution :**

In the question given that,

A line passing through

(1, -2)Let us considered that, the equation of the line cutting

equal intercepts at coordinates of length â€˜mâ€™ is

According to the formula of the equation of the line

x/a + y/b = 1

We get

x/m + y/m = 1

x + y = m

The straight line x + y = m passes through (1, -2)

So, the given point satisfy the equation

1 – 2 = m

m = -1

Hence, the equation of the line is x+ y = -1

### Question 3. Find the equation to the straight line which passes through the point (5, 6) and has intercepts on the axes

### (1) Equal in magnitude and both positive

**Solution:**

We have given that,

a = b

Now find the equation of line cutoff intercepts from the axes:

According to the formula of the equation of the line

x/a + y/b = 1

We get

x + y = a

It is given that the straight line which is passes through the point

(5, 6)So, the given point satisfy the equation

5 + 6 = a

a = 11

Hence, the equation of the line is

x + y = 11

### (2) Equal in magnitude but opposite in sign

**Solution:**

We have given that,

b = -a

Now we find the equation of line cutoff intercepts from the axes:-

According to the formula of the equation of the line

x/a + y/b = 1

We get

x/a + y/-a = 1

x â€“ y = a

It is given that the straight line which is passes through the point

(5, 6)So, the given point satisfy the equation

5 â€“ 6 = a

a = -1

Hence, the equation of the line is x â€“ y = -1

### Question 4. For what values of a and b the intercepts cut off on the coordinate axes by the line ax + by + 8 = 0 are equal in length but opposite in signs to those cut off by the line 2x â€“ 3y + 6 = 0 on the axes.

**Solution:**

According to the question

It is given that intercepts cut off on the coordinate axes by the line

ax + by +8 = 0 …(1)

So the slope of the equations is

slope(m1) = â€“a/b

Also, they are equal in length, but opposite in signs to those cut off by the line

2x â€“ 3y +6 = 0 …(2)

So the slope of the equations is

slope(m2) = 2/3

So, on solving we get

-a/b = 2/3

a = -2b/3 …..(3)

Now, the length of the perpendicular from the origin.

So, by using the formula, we get

d = |ax + by + c/âˆša

^{2 }+ b^{2}|d

_{1 }= |a(0) + b(0) + 8/âˆša^{2 }+ b^{2}|= 8 Ã— 3/âˆš13b

^{2}d

_{2 }= |2(0) – 3(0) + 6/âˆš2^{2 }+ 3^{2}|= 6/âˆš13

As we know that d

_{1}= d_{2},So, 8 Ã— 3/âˆš13b

^{2 }= 6/âˆš13b = 4

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

a = -2b/3

a = -8/3

Hence, the value of a and b is -8/3 and 4

### Question 5. Find the equation to the straight line which cuts off equal positive intercepts on the axes and their product is 25.

**Solution:**

According to the question

a = b …..(1)

ab = 25 …..(2)

We have to find the equation of the line which cutoff equal positive intercepts on the axes

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

a

^{2}= 25a = 5

According to the formula of the equation of the line

x/a + y/b = 1

We get

x/5 + y/5 = 1

Now we’re taking LCM, we get

x + y = 5

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

### Question 6. Find the equation of the line which passing through the point (-4, 3), and the portion of the line intercepted between the axes is divided internally in the ratio 5:3 by this point.

**Solution:**

As we know that the equation of the line is

x/a + y/b = 1 …..(1)

And it cuts the axis at point A(a, 0) and B(0, b)

So, AB intercept between the axis is 5 : 3

h = 3 Ã— a + 5 Ã— 0/8

k = 3 Ã— 0 + 5 Ã— 0/8

P = (3a/8, 5b/8)

It is given that the straight line which is passes through the point

(-4, 3)So, 3a/8 = -4 and 5b/8 = 3

a = -32/3 and 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 line is 9x – 20y + 96 = 0

### Question 7. A straight line passes through the point bisect the portion of the line intercepted between the axes. show that the eq of line is x/2Î± + y/2Î² = 1.

**Solution:**

As we know that the equation of the line is

x/a + y/b = 1 …..(1)

And line intercept by the axes are(a, 0) and (0, b),

If the line segment bisect at the point (Î±, Î²) then,

a + 0/2 = Î±

0 + b/2 = Î²

a = 2Î±, b = 2Î²

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

x/2Î± + y/Î² = 1

### Question 8. Find the equation of line which passes through the point (3, 4) and is such that are the portion of it intercepted between the axes is divisible by the point in the ratio 2:3.

**Solution:**

As we know that the equation of the line is

x/a + y/b = 1 …..(1)

And let us assume point Q(3, 4) divides the line that point A(a, 0) and B(0, b)

in ratio 2 : 3

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

2(a) + 3(0)/2 + 3 = 4

a = 5, b = 10

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

x/5 + y/10 =1

Hence, the equation of line is 2x + y = 10

### Question 9. Point R(h, k) divides a line segment between the axes in the ratio 1 : 2. Find the equation of line.

**Solution:**

It is given that the point R(h, k) divides a line segment between the axes in the ratio 1 : 2

By using the section formula we get

h = 2 Ã— a + 1 Ã— 0/1 + 2

and,

k = 2 Ã— 0 + 1 Ã— b/1 + 2

So,

h = 2a/3 and k = b/3

Hence, the value of a and b is

a = 3h/2

b = 3k

Thus, the corresponding points of A (3h/2, 0) and B (0, 3k)

y – 3k/3k – 0 = x – 0/0 – 3h/2

-3hy + 9hk = 6kx

Hence, the equation of the line is 2kx + hy = 3kh

### Question 10. Find the equation of the line which passes through the point (-3, 8) and cuts off positive intercepts on the coordinates axes whose sum is 7.

**Solution:**

As we know that the equation of the line is

x/a + y/b = 1 …..(1)

Then a + b = 7

x/a + y/7 – 1 = 1 …..(2)

It is given that the line passes through point(-3, 8)

= -3/a + 8/7 – a = 1

=-21 + 3a + 8a = 7a – a

^{2}= a

^{2 }+ 4a – 21 = 0a = 3 or -7

But a > 0 so a not = -7

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

x/3 + y/4 = 1

Hence, the equation of the line is 4x + 3y = 12

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