# NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.3 Straight Lines

#### NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.3 Straight Lines

The NCERT Solutions for Class 11 Maths Chapter 9, covering Straight Lines, are a helpful resource for students preparing for their Math board exams. The chapter covers important topics like slope, Horizontal and vertical Lines, Point-slope form, Two-point form, Slope-intercept form, Intercept – form, and Normal form. These solutions are designed to help students practice and improve their understanding using straightforward methods and formulas.

### NCERT Solutions for Class 11 Maths Chapter 09

#### EXERCISE 9.3 PAGE NO: 167

**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) **x + 7y = 0

Given:

The equation is x + 7y = 0

The slope-intercept form is represented in the form ‘y = mx + c’, where m is the slope and c is the y-intercept.

So, the above equation can be expressed as

y = -1/7x + 0

∴ The above equation is of the form y = mx + c, where m = -1/7 and c = 0

**(ii) **6x + 3y – 5 = 0

Given:

The equation is 6x + 3y – 5 = 0

The slope-intercept form is represented in the form ‘y = mx + c’, where m is the slope and c is the y-intercept.

So, the above equation can be expressed as

3y = -6x + 5

y = -6/3x + 5/3

= -2x + 5/3

∴ The above equation is of the form y = mx + c, where m = -2 and c = 5/3

**(iii) **y = 0

Given:

The equation is y = 0

The slope-intercept form is given by ‘y = mx + c’, where m is the slope and c is the y-intercept.

y = 0 × x + 0

∴ The above equation is of the form y = mx + c, where m = 0 and c = 0

**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) **3x + 2y – 12 = 0

Given:

The equation is 3x + 2y – 12 = 0

The equation of the line in intercept form is given by x/a + y/b = 1, where ‘a’ and ‘b’ are intercepted on the x-axis and the y-axis, respectively.

So, 3x + 2y = 12

Now, let us divide both sides by 12; we get

3x/12 + 2y/12 = 12/12

x/4 + y/6 = 1

∴ The above equation is of the form x/a + y/b = 1, where a = 4, b = 6

The intercept on the x-axis is 4.

The intercept on the y-axis is 6.

**(ii) **4x – 3y = 6

Given:

The equation is 4x – 3y = 6

The equation of the line in intercept form is given by x/a + y/b = 1, where ‘a’ and ‘b’ are intercepted on the x-axis and the y-axis, respectively.

So, 4x – 3y = 6

Now, let us divide both sides by 6; we get

4x/6 – 3y/6 = 6/6

2x/3 – y/2 = 1

x/(3/2) + y/(-2) = 1

∴ The above equation is of the form x/a + y/b = 1, where a = 3/2, b = -2

The intercept on the x-axis is 3/2.

The intercept on the y-axis is -2.

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

Given:

The equation is 3y + 2 = 0

The equation of the line in intercept form is given by x/a + y/b = 1, where ‘a’ and ‘b’ are intercepted on the x-axis and the y-axis, respectively.

So, 3y = -2

Now, let us divide both sides by -2; we get

3y/-2 = -2/-2

3y/-2 = 1

y/(-2/3) = 1

∴ The above equation is of the form x/a + y/b = 1, where a = 0, b = -2/3

The intercept on the x-axis is 0.

The intercept on the y-axis is -2/3.

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

**Solution:**

Given:

The equation of the line is 12(x + 6) = 5(y – 2).

12x + 72 = 5y – 10

12x – 5y + 82 = 0 … (1)

Now, compare equation (1) with the general equation of line Ax + By + C = 0, where A = 12, B = –5, and C = 82

Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x_{1}, y_{1}) is given by

∴ The distance is 5 units.

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

**Solution:**

Given:

The equation of the line is x/3 + y/4 = 1

4x + 3y = 12

4x + 3y – 12 = 0 …. (1)

Now, compare equation (1) with the general equation of line Ax + By + C = 0, where A = 4, B = 3, and C = -12

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

So, the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x_{1}, y_{1}) is given by

|4a – 12| = 4 × 5

**±** (4a – 12) = 20

4a – 12 = 20 or – (4a – 12) = 20

4a = 20 + 12 or 4a = -20 + 12

a = 32/4 or a = -8/4

a = 8 or a = -2

∴ The required points on the x-axis are (-2, 0) and (8, 0)

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:**

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

Given:

The parallel lines are 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0.

By using the formula,

The distance (d) between parallel lines Ax + By + C_{1} = 0 and Ax + By + C_{2} = 0 is given by

∴ The distance between parallel lines is 65/17

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

Given:

The parallel lines are l (x + y) + p = 0 and l (x + y) – r = 0

lx + ly + p = 0 and lx + ly – r = 0

By using the formula,

The distance (d) between parallel lines Ax + By + C_{1} = 0 and Ax + By + C_{2} = 0 is given by

∴ The distance between parallel lines is |p+r|/l**√**2

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

**Solution:**

Given:

The line is 3x – 4y + 2 = 0

So, y = 3x/4 + 2/4

= 3x/4 + ½

Which is of the form y = mx + c, where m is the slope of the given line.

The slope of the given line is 3/4

We know that parallel lines have the same slope.

∴ Slope of other line = m = 3/4

The equation of line having slope m and passing through (x_{1}, y_{1}) is given by

y – y_{1} = m (x – x_{1})

∴ The equation of the line having slope 3/4 and passing through (-2, 3) is

y – 3 = ¾ (x – (-2))

4y – 3 × 4 = 3x + 3 × 2

3x – 4y = 18

∴ The equation is 3x – 4y = 18

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

**Solution:**

Given:

The equation of line is x – 7y + 5 = 0

So, y = 1/7x + 5/7 [which is of the form y = mx + c, where m is the slope of the given line.]

The slope of the given line is 1/7

The slope of the line perpendicular to the line having slope m is -1/m

The slope of the line perpendicular to the line having a slope of 1/7 is -1/(1/7) = -7

So, the equation of the line with slope -7 and the x-intercept 3 is given by y = m(x – d)

y = -7 (x – 3)

y = -7x + 21

7x + y = 21

∴ The equation is 7x + y = 21

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

**Solution:** Given:

The lines are √3x + y = 1 and x + √3y = 1

So, y = -√3x + 1 … (1) and

y = -1/√3x + 1/√3 …. (2)

The slope of the line (1) is m_{1} = -√3, while the slope of the line (2) is m_{2} = -1/√3

Let θ be the angle between two lines.

So,

θ = 30°

∴ The angle between the given lines is either 30° or 180°- 30° = 150°

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

**Solution:**

Let the slope of the line passing through (h, 3) and (4, 1) be m_{1}

Then, m_{1} = (1-3)/(4-h) = -2/(4-h)

Let the slope of line 7x – 9y – 19 = 0 be m_{2}

7x – 9y – 19 = 0

So, y = 7/9x – 19/9

m_{2} = 7/9

Since the given lines are perpendicular,

m_{1} × m_{2} = -1

-2/(4-h) × 7/9 = -1

-14/(36-9h) = -1

-14 = -1 × (36 – 9h)

36 – 9h = 14

9h = 36 – 14

h = 22/9

∴ The value of h is 22/9

**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:**

Let the slope of line Ax + By + C = 0 be m

Ax + By + C = 0

So, y = -A/Bx – C/B

m = -A/B

By using the formula,

Equation of the line passing through point (x_{1}, y_{1}) and having slope m = -A/B is

y – y_{1} = m (x – x_{1})

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

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

Hence, proved.

**11. Two lines passing through point (2, 3) intersects each other at an angle of 60 ^{o}. If the slope of one line is 2, find the equation of the other line.**

**Solution:**

Given: m_{1} = 2

Let the slope of the first line be m_{1}

And let the slope of the other line be m_{2}.

The angle between the two lines is 60°.

So,

**12. Find the equation of the right bisector of the line segment joining the points (3, 4) and (–1, 2).**

**Solution:**

Given:

The right bisector of a line segment bisects the line segment at 90°.

End-points of the line segment AB are given as A (3, 4) and B (–1, 2).

Let the mid-point of AB be (x, y).

x = (3-1)/2= 2/2 = 1

y = (4+2)/2 = 6/2 = 3

(x, y) = (1, 3)

Let the slope of line AB be m_{1}

m_{1} = (2 – 4)/(-1 – 3)

= -2/(-4)

= 1/2

And let the slope of the line perpendicular to AB be m_{2}

m_{2} = -1/(1/2)

= -2

The equation of the line passing through (1, 3) and having a slope of –2 is

(y – 3) = -2 (x – 1)

y – 3 = – 2x + 2

2x + y = 5

∴ The required equation of the line is 2x + y = 5

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

**Solution:**

Let us consider the coordinates of the foot of the perpendicular from (-1, 3) to the line 3x – 4y – 16 = 0 be (a, b)

So, let the slope of the line joining (-1, 3) and (a, b) be m_{1}

m_{1 }= (b-3)/(a+1)

And let the slope of the line 3x – 4y – 16 = 0 be m_{2}

y = 3/4x – 4

m_{2} = 3/4

Since these two lines are perpendicular, m_{1} × m_{2} = -1

(b-3)/(a+1) × (3/4) = -1

(3b-9)/(4a+4) = -1

3b – 9 = -4a – 4

4a + 3b = 5 …….(1)

Point (a, b) lies on the line 3x – 4y = 16

3a – 4b = 16 ……..(2)

Solving equations (1) and (2), we get

a = 68/25 and b = -49/25

∴ The coordinates of the foot of perpendicular are (68/25, -49/25)

**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:**

Given:

The perpendicular from the origin meets the given line at (–1, 2).

The equation of the line is y = mx + c

The line joining the points (0, 0) and (–1, 2) is perpendicular to the given line.

So, the slope of the line joining (0, 0) and (–1, 2) = 2/(-1) = -2

The slope of the given line is m.

m × (-2) = -1

m = 1/2

Since point (-1, 2) lies on the given line,

y = mx + c

2 = 1/2 × (-1) + c

c = 2 + 1/2 = 5/2

∴ The values of m and c are 1/2 and 5/2, respectively.

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

**Solution:**

Given:

The equations of the given lines are

x cos θ – y sin θ = k cos 2θ …………………… (1)

x sec θ + y cosec θ = k ……………….… (2)

Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x_{1}, y_{1}) is given by

q = k cos θ sin θ

Multiply both sides by 2, and we get

2q = 2k cos θ sin θ = k × 2sin θ cos θ

2q = k sin 2θ

Squaring both sides, we get

4q^{2} = k^{2} sin^{2}2θ …………………(4)

Now add (3) and (4); we get

p^{2} + 4q^{2} = k^{2} cos^{2} 2θ + k^{2} sin^{2} 2θ

p^{2} + 4q^{2} = k^{2} (cos^{2} 2θ + sin^{2} 2θ) [Since, cos^{2} 2θ + sin^{2} 2θ = 1]

∴ p^{2} + 4q^{2} = k^{2}

Hence proved.

**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 vertex A.**

**Solution:**

Let AD be the altitude of triangle ABC from vertex A.

So, AD is perpendicular to BC.

Given:

Vertices A (2, 3), B (4, –1) and C (1, 2)

Let the slope of the line BC = m_{1}

m_{1} = (- 1 – 2)/(4 – 1)

m_{1} = -1

Let the slope of the line AD be m_{2}

AD is perpendicular to BC.

m_{1} × m_{2} = -1

-1 × m_{2} = -1

m_{2} = 1

The equation of the line passing through the point (2, 3) and having a slope of 1 is

y – 3 = 1 × (x – 2)

y – 3 = x – 2

y – x = 1

Equation of the altitude from vertex A = y – x = 1

Length of AD = Length of the perpendicular from A (2, 3) to BC

The equation of BC is

y + 1 = -1 × (x – 4)

y + 1 = -x + 4

x + y – 3 = 0 …………………(1)

Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x_{1}, y_{1}) is given by

Now, compare equation (1) to the general equation of the line, i.e., Ax + By + C = 0; we get

Length of AD =

[where, A = 1, B = 1 and C = -3]

∴ The equation and the length of the altitude from vertex A are y – x = 1 and

√2 units, respectively.

**17. If p is the length of the 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:**

The equation of a line whose intercepts on the axes are a and b is x/a + y/b = 1

bx + ay = ab

bx + ay – ab = 0 ………………..(1)

Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x_{1}, y_{1}) is given by

Now, square on both sides; we get

∴ 1/p^{2} = 1/a^{2} + 1/b^{2}

Hence, proved.