**NCERT Solutions for Class 11 Maths Chapter 9 ****Straight Lines**** Ex 9.3**

**Ex 9.3 Class 11 Maths Question 1.**

### Reduce the following equations into slope - intercept form and find their slopes and the y-intercepts.

**(i)** x + 7y = 0,

**(ii)** 6x + 3y – 5 = 0,

**(iii)** y = 0

**Solution:****(i)** The given equation is x + 7y = 0, which can be written in the form

7y = –x ⇒ y = −x/7 + 0 … (1)

Also, the slope - intercept form is y = mx + c …(2)

On comparing (1) and (2), we get

m = −1/7, c = 0

Hence, the slope is −1/7 and the y-intercept = 0.

**(ii)** The given equation is 6x + 3y – 5 = 0, which can be written in the form

3y = –6x + 5

⇒ y = –2x + 5/3 …(1)

Also, the slope - intercept form is y = mx + c … (2)

On comparing (1) and (2), we get

m = –2 and c = 5/3

i.e. slope = –2 and the y-intercept = 5/3

**(iii)** The given equation is y = 0, which can be written in the form

y = 0·x + 0 … (1)

Also, the slope - intercept form is y = mx + c … (2)

On comparing (1) and (2), we get

m = 0, c = 0.

Hence, slope is 0 and the y-intercept is 0.

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**Ex 9.3 Class 11 Maths Question 2.**

### Reduce the following equations into intercept form and find their intercepts on the axes.

**(i)** 3x + 2y – 12 = 0,

**(ii)** 4x – 3y = 6,

**(iii)** 3y + 2 = 0

**Solution:****(i)** The given equation is 3x + 2y – 12 = 0

We have to reduce the given equation into intercept form,

i.e., x/a + y/b = 1 …(1)

Now, the given equation can be written as 3x + 2y = 12

⇒ 3x/12 + 2y/12 = 1 ⇒ x/4 + y/6 = 1 …(2)

On comparing (1) and (2), we get a = 4, b = 6

Hence, the intercepts of the line are 4 and 6.

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**(ii)** The given equation is 4x – 3y = 6

We have to reduce the given equation into intercept form,

i.e., x/a + y/b = 1 …(1)

Now, the given equation can be written as

4x/6 − 3y/6 = 1 or 2x/3 – y/2 = 1 …(2)

On comparing (1) and (2), we get

a = 3/2 and b = –2

Hence, the intercepts of the line are 3/2 and –2.

**(iii)** The given equation is 3y + 2 = 0

We have to reduce the given equation into intercept form,

i.e., x/a + y/b = 1 …(1)

Now, the given equation can be written as

### 3y = -2

0.x/-2 + 3y/-2 = 1 or 0.x/-2 + y/(-2/3) = 1 …(2)

On comparing (1) and (2), we get

a = 0 and b = –2/3

Hence, the intercepts of the line are 0 and –2/3.

**Ex 9.3 Class 11 Maths Question 3.**

### Find the distance of the point (-1, 1) from the line 12(x + 6) = 5(y – 2).

**Solution:**

**Ex 9.3 Class 11 Maths Question 4.**

### Find the points on the x-axis, whose distances from the line x/3 + y/4 = 1 are 4 units.

**Solution:**

The given equation of line is x/3 + y/4 = 1, which can be written as

4x + 3y – 12 = 0 … (i)

Let (a, 0) be the point on x-axis whose distance from line (i) is 4 units.

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**Ex 9.3 Class 11 Maths Question 5.**

### Find the distance between parallel lines

**(i)** 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0

**(ii)** *l*(x + y) + p = 0 and *l*(x + y) – r = 0.

**Solution:**

If the lines are Ax + By + C_{1} = 0 and Ax + By + C_{2} = 0, then distance between

**Ex 9.3 Class 11 Maths Question 6.**

### Find equation of the line parallel to the line 3x – 4y + 2 = 0 and passing through the point (-2, 3).

**Solution:**

The given equation of line is 3x – 4y + 2 = 0 … (i)

This equation can be written as y = 3x/4 + ½

Slop of the line = 3/4

Thus, slope of any line parallel to the given line (i) is 3/4 and passes through (-2, 3), then its equation is

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**Ex 9.3 Class 11 Maths Question 7.**

### Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having x intercept 3.

**Solution:**

The given equation of line is x – 7y + 5 = 0 … (i)

This equation can be written as y = x/7 + 5/7

Slope of this line = 1/7

Thus, slope of any line perpendicular to the line (i) is -7 and passes through (3, 0), then its equation is

y – 0 = -7(x – 3) [∵ Product of slope of perpendicular lines is -1]

⇒ y = -7x + 21

⇒ 7x + y – 21 = 0

This is the required equation of line.

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**Ex 9.3 Class 11 Maths Question 8.**

### Find angles between the lines √3x + y = 1 and x + √3y = 1.

**Solution:**

The given equations are

√3x + y = 1 … (i)

x + √3y = 1 … (ii)

Since we have to find the angle between the two lines, therefore, firstly we need to find the slopes of lines (i) and (ii).

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**Ex 9.3 Class 11 Maths Question 9.**

### The line through the points (h, 3) and (4, 1) intersects the line 7x – 9y – 19 = 0 at right angle. Find the value of h.

**Solution:**

The given points are (h, 3) and (4, 1).

∴ Slope of the line joining the points (h, 3) and (4, 1)

**Ex 9.3 Class 11 Maths Question 10.**

### Prove that the line through the point (x_{1}, y_{1}) and parallel to the line Ax + By + C = 0 is A(x – x_{1}) + B(y – y_{1}) = 0.

**Solution:**

The given equation of line is Ax + By + C = 0

∴ Slope of the above line = −A/B

i.e. slope of any line parallel to given line and passing through (x_{1}, y_{1}) is −A/B.

Then, equation is (y – y_{1}) = −A/B (x – x_{1})

⇒ B(y – y_{1}) = -A(x – x_{1})

⇒ A(x – x_{1}) + B(y – y_{1}) = 0

Hence proved.

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**Ex 9.3 Class 11 Maths Question 11.**

### Two lines passing through the point (2, 3) intersects each other at an angle of 60°. If slope of one line is 2, find equation of the other line.

**Solution:**

We have given that two lines are passing through the point (2, 3) and intersects each other at an angle of 60°.

Let m be the slope of the other line.

**Ex 9.3 Class 11 Maths Question 12.**

### Find the equation of the right bisector of the line segment joining the points (3, 4) and (-1, 2).

**Solution:**

Let the given points are A(3, 4) and B(-1, 2).

Let M be the mid-point of AB.

**Ex 9.3 Class 11 Maths Question 13.**

### Find the coordinates of the foot of perpendicular from the point (-1, 3) to the line 3x – 4y – 16 = 0.

**Solution:**

The given line is 3x – 4y – 16 = 0 … (i)

Slope of the line (i) = 3/4

Then, equation of any line perpendicular from (-1, 3) to the given line (i) is:

**Ex 9.3 Class 11 Maths Question 14.**

### The perpendicular from the origin to the line y = mx + c meets it at the point (-1, 2). Find the values of m and c.

**Solution:**

We have given that the perpendicular from the origin to the line y = mx + c meets it at the point (-1, 2).

∴ 2 = m(-1) + c … (i)

⇒ c – m = 2

**Ex 9.3 Class 11 Maths Question 15.**

### If p and q are the lengths of perpendiculars from the origin to the lines x cos Î¸ – y sin Î¸ = k cos 2Î¸ and x sec Î¸ + y cosec Î¸ = k, respectively, prove that p^{2} + 4q^{2} = k^{2}.

**Solution:**

We have given that p and q are the lengths of perpendiculars from the origin to the lines x cos Î¸ – y sin Î¸ = k cos 2Î¸ and x sec Î¸ + y cosec Î¸ = k.

**Ex 9.3 Class 11 Maths Question 16.**

### In the triangle ABC with vertices A(2, 3), B(4, -1) and C( 1, 2), find the equation and length of altitude from the vertex A.

**Solution:**

We have given a AABC with the vertices, A (2, 3), B (4, -1) and C (1, 2)

**Ex 9.3 Class 11 Maths Question 17.**

### If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b, then show that 1/p^{2} = 1/a^{2} + 1/b^{2}.

**Solution:**

We have given that p is the length of perpendicular from the origin to the line whose intercepts

**Related Links:**

**NCERT Solutions for Maths Class 9**