NCERT Solutions are provided to help the students in understanding the steps to solve mathematical problems that are provided in the textbook. The Exercise 10.3 of NCERT Solutions for Class 11 Maths Chapter 10- Straight Lines is based on the following topics:

- General Equation of a Line
- Different forms of Ax + By + C = 0

- Distance of a Point From a Line
- Distance between two parallel lines

The subject experts at BYJUâ€™S stick to the syllabus while preparing the solutions. The problem-solving method provided in the examples is followed while preparing the NCERT Solutions for class 11 as well.

## Download PDF of NCERT Solutions for Class 11 Maths Chapter 10- Straight Lines Exercise 10.3

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### Access other exercise solutions of Class 11 Maths Chapter 10

Exercise 10.1 Solutions 14 Questions

Exercise 10.2 Solutions 20 Questions

Miscellaneous Exercise On Chapter 10 Solutions 24 Questions

#### Access Solutions for Class 11 Maths Chapter 10.3 Exercise

**1. Reduce the following equations into slope â€“ intercept form and find their slopes and the y â€“ intercepts.(i) x + 7y = 0**

**(ii) 6x + 3y â€“ 5 = 0(iii) y = 0**

**Solution:**

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

Given:

The equation is x + 7y = 0

Slope â€“ intercept form is represented in the form â€˜y = mx + câ€™, where m is the slope and c is the y intercept

So, the above equation can be expressed as

y = -1/7x + 0

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

**(ii) **6x + 3y â€“ 5 = 0

Given:

The equation is 6x + 3y â€“ 5 = 0

Slope â€“ intercept form is represented in the form â€˜y = mx + câ€™, 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

âˆ´ 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

Slope â€“ intercept form is given by â€˜y = mx + câ€™, where m is the slope and c is the y intercept

y = 0 Ã— x + 0

âˆ´ 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 â€“ 12 = 0**

**(ii) 4x â€“ 3y = 6**

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

**Solution:**

**(i) **3x + 2y â€“ 12 = 0

Given:

The equation is 3x + 2y â€“ 12 = 0

Equation of line in intercept form is given by x/a + y/b = 1, where â€˜aâ€™ and â€˜bâ€™ are intercepts on x axis and y â€“ 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

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

Intercept on x â€“ axis is 4

Intercept on y â€“ axis is 6

**(ii) **4x â€“ 3y = 6

Given:

The equation is 4x â€“ 3y = 6

Equation of line in intercept form is given by x/a + y/b = 1, where â€˜aâ€™ and â€˜bâ€™ are intercepts on x axis and y â€“ axis respectively.

So, 4x â€“ 3y = 6

Now let us divide both sides by 6, we get

4x/6 â€“ 3y/6 = 6/6

2x/3 â€“ y/2 = 1

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

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

Intercept on x â€“ axis is 3/2

Intercept on y â€“ axis is -2

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

Given:

The equation is 3y + 2 = 0

Equation of line in intercept form is given by x/a + y/b = 1, where â€˜aâ€™ and â€˜bâ€™ are intercepts on x axis and y â€“ 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

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

Intercept on x â€“ axis is 0

Intercept on y â€“ axis is -2/3

**3. Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive x-axis.**

**(i) x â€“ âˆš3y + 8 = 0**

**(ii) y â€“ 2 = 0**

**(iii) x â€“ y = 4**

**Solution:**

**(i) **x â€“ âˆš3y + 8 = 0

Given:

The equation is x â€“ âˆš3y + 8 = 0

Equation of line in normal form is given by x cos Î¸ + y sin Î¸ = p where â€˜Î¸â€™ is the angle between perpendicular and positive x axis and â€˜pâ€™ is perpendicular distance from origin.

So now, x â€“ âˆš3y + 8 = 0

x â€“ âˆš3y = -8

Divide both the sides by âˆš(1^{2} + (âˆš3)^{2}) = âˆš(1 + 3) = âˆš4 = 2

x/2 â€“ âˆš3y/2 = -8/2

(-1/2)x + âˆš3/2y = 4

This is in the form of: x cos 120^{o} + y sin 120^{o} = 4

âˆ´ The above equation is of the form x cos Î¸ + y sin Î¸ = p, where Î¸ = 120Â° and p = 4.

Perpendicular distance of line from origin = 4

Angle between perpendicular and positive x â€“ axis = 120Â°

**(ii) **y â€“ 2 = 0

Given:

The equation is y â€“ 2 = 0

Equation of line in normal form is given by x cos Î¸ + y sin Î¸ = p where â€˜Î¸â€™ is the angle between perpendicular and positive x axis and â€˜pâ€™ is perpendicular distance from origin.

So now, 0 Ã— x + 1 Ã— y = 2

Divide both sides by âˆš(0^{2} + 1^{2}) = âˆš1 = 1

0 (x) + 1 (y) = 2

This is in the form of: x cos 90^{o} + y sin 90^{o} = 2

âˆ´ The above equation is of the form x cos Î¸ + y sin Î¸ = p, where Î¸ = 90Â° and p = 2.

Perpendicular distance of line from origin = 2

Angle between perpendicular and positive x â€“ axis = 90Â°

**(iii) **x â€“ y = 4

Given:

The equation is x â€“Â y + 4 = 0

Equation of line in normal form is given by x cos Î¸ + y sin Î¸ = p where â€˜Î¸â€™ is the angle between perpendicular and positive x axis and â€˜pâ€™ is perpendicular distance from origin.

So now, x â€“ y = 4

Divide both the sides by âˆš(1^{2} + 1^{2}) = âˆš(1+1) = âˆš2

x/âˆš2 â€“ y/âˆš2 = 4/âˆš2

1/âˆš2x + (-1/âˆš2)y = 2âˆš2

This is in the form: x cos 315^{o} + y sin 315^{o} = 2âˆš2

âˆ´ The above equation is of the form x cos Î¸ + y sin Î¸ = p, where Î¸ = 315Â° and p = 2âˆš2.

Perpendicular distance of line from origin = 2âˆš2

Angle between perpendicular and positive x â€“ axis = 315Â°

**4. Find the distance of the point (â€“1, 1) from the line 12(x + 6) = 5(y â€“ 2).**

**Solution:**

Given:

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

12x + 72 = 5y â€“ 10

12x â€“ 5y + 82 = 0 â€¦ (1)

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

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

âˆ´ The distance is 5units.

**5. 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 line isÂ x/3 + y/4 = 1

4x + 3y = 12

4x + 3y â€“ 12 = 0 â€¦. (1)

Now, compare equation (1) with 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 â€“ 12| = 4 Ã— 5

Â **Â±** (4a â€“ 12) = 20

4a â€“ 12 = 20 or â€“ (4a â€“ 12) = 20

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

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

a = 8 or a = -2

âˆ´Â The required points on the x â€“ axis are (-2, 0) and (8, 0)

**6. Find the distance between parallel lines (i) 15x + 8y â€“ 34 = 0 and 15x + 8y + 31 = 0**

**(ii) l(x + y) + p = 0 and l (x + y) â€“ r = 0**

**Solution:**

**(i) **15x + 8y â€“ 34 = 0 and 15x + 8y + 31 = 0

Given:

The parallel lines are 15x + 8y â€“ 34 = 0 and 15x + 8y + 31 = 0.

By using the formula,

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

= 65/**âˆš**289

= 65/17

âˆ´Â The distance between parallel lines is 65/17

**(ii) **l(x + y) + p = 0 and l (x + y) â€“ r = 0

Given:

The parallel lines are l (x + y) + p = 0 and l (x + y) â€“ r = 0.

lx + ly + p = 0 and lx + ly â€“ r = 0

by using the formula,

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

âˆ´Â The distance between parallel lines is |p+q|/l**âˆš**2

**7. Find equation of the line parallel to the line 3x âˆ’ 4y + 2 = 0 and passing through the point (â€“2, 3).**

**Solution:**

Given:

The line is 3x â€“ 4y + 2 = 0

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

= 3x/4 + Â½

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 line have same slope.

âˆ´Â Slope of other line = m = 3/4

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

y â€“ y_{1} = m (x â€“ x_{1})

âˆ´Â Equation of line having slope 3/4 and passing through (-2, 3) is

y â€“ 3 = Â¾ (x â€“ (-2))

4y â€“ 3 Ã— 4 = 3x + 3 Ã— 2

3x â€“ 4y = 18

âˆ´Â The equation is 3x â€“ 4y = 18

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

**Solution:**

Given:

The equation of line is x â€“ 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.]

Slope of the given line is 1/7

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

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

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

y = -7 (x â€“ 3)

y = -7x + 21

7x + y = 21

âˆ´Â The equation is 7x + y = 21

**9. Find angles between the lines âˆš3x + y = 1 andÂ x +Â âˆš3y = 1.**

**Solution:**

Given:

The lines are âˆš3x + y = 1 andÂ x +Â âˆš3y = 1

So, y = -âˆš3x + 1 â€¦ (1) and

y = -1/âˆš3x + 1/âˆš3 â€¦. (2)

Slope of line (1) is m_{1}Â = -âˆš3, while the slope of line (2) is m_{2}Â = -1/âˆš3

Let Î¸ be the angle between two lines

So,

Î¸ = 30Â°

âˆ´ The angle between the given lines is either 30Â° or 180Â°- 30Â° = 150Â°

**10. The line through the points (h, 3) and (4, 1) intersects the line 7x âˆ’ 9y âˆ’19 = 0. At 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)Â

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

7x â€“ 9y â€“ 19 = 0

So, y = 7/9x â€“ 19/9

m_{2}Â = 7/9

Since, the given lines are perpendicular

m_{1}Â Ã— m_{2}Â = -1

-2/(4-h)Â Ã— 7/9 = -1

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

-14 = -1 Ã— (36 â€“ 9h)

36 â€“ 9h = 14

9h = 36 â€“ 14

h = 22/9

âˆ´ The value of h is 22/9

**11. Prove that the line through the point (x _{1}, y_{1}) and parallel to the line Ax + By + C = 0 is A (x â€“ x_{1}) + B (y â€“ y_{1}) = 0.**

**Solution:**

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

Ax + By + C = 0

So, y = -A/Bx â€“ 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 â€“ y_{1} = m (x â€“ x_{1})

= -A/B (x â€“ x_{1})

B (y â€“ y_{1}) = -A (x â€“ x_{1})

âˆ´Â A(x â€“ x_{1}) + B(y â€“ y_{1}) = 0

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

Hence proved.

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

**Solution:**

Given: m_{1}Â = 2

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

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

Angle between the two lines is 60Â°.

So,

**13. Find the equation of the right bisector of the line segment joining the points (3, 4) and (â€“1, 2).**

**Solution:**

Given:

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

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

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

x = (3+4)y/2

y = (-1+2)/2

(x, y) = (7/2, 1/2)

Let the slope of line AB be m_{1}

m_{1}Â = (2 â€“ 4)/(-1 â€“ 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 â€“2 is

(y â€“ 3) = -2 (x â€“ 1)

y â€“ 3 = â€“ 2x + 2

2x + y = 5

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

**14. Find the coordinates of the foot of perpendicular from the point (â€“1, 3) to the line 3x â€“ 4y â€“ 16 = 0.**

**Solution:**

Let us consider the co-ordinates of the foot of the perpendicular from (-1, 3) to the line 3x â€“ 4y â€“ 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 â€“ 4y â€“ 16 = 0 be m_{2}

y = 3/4x â€“ 4

m_{2} = 3/4

Since these two lines are perpendicular, m_{1}Â Ã— m_{2}Â = -1

(b-3)/(a+1) Ã— (3/4) = -1

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

3b â€“ 9 = -4a â€“ 4

4a + 3b = 5 â€¦â€¦.(1)

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

3a â€“ 4b = 16 â€¦â€¦..(2)

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

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

âˆ´ The co-ordinates of the foot of perpendicular is (68/25, -49/25)

**15. The perpendicular from the origin to the line y = mx + c meets it at the point (â€“1, 2). Find the values of m and c.**

**Solution:**

Given:

The perpendicular from the origin meets the given line at (â€“1, 2).

The equation of line is y = mx + c

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

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

Slope of the given line is m.

m Ã— (-2) = -1

m = 1/2

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

y = mx + c

2 = 1/2 Ã— (-1) + c

c = 2 + 1/2 = 5/2

âˆ´ The values of m and c are 1/2 and 5/2 respectively.

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

**Solution:**

Given:

The equations of given lines are

x cos Î¸ â€“ y sin Î¸ = k cos 2Î¸ â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (1)

x sec Î¸ + y cosec Î¸ = k â€¦â€¦â€¦â€¦â€¦â€¦.â€¦ (2)

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

q = k cos Î¸ sin Î¸

Multiply both sides by 2, we get

2q = 2k cos Î¸ sin Î¸ = k Ã— 2sin Î¸ cos Î¸

2q = k sin 2Î¸

Squaring both sides, we get

4q^{2}Â = k^{2}Â sin^{2}2Î¸ â€¦â€¦â€¦â€¦â€¦â€¦â€¦(4)

Now add (3) and (4) we get

p^{2}Â + 4q^{2}Â = k^{2}Â cos^{2}Â 2Î¸ + k^{2}Â sin^{2}Â 2Î¸

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

âˆ´Â p^{2}Â + 4q^{2}Â = k^{2}

Hence proved.

**17. In the triangle ABC with vertices A (2, 3), B (4, â€“1) and C (1, 2), find the equation and length of altitude from the 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, â€“1) and C (1, 2)

Let slope of line BC = m_{1}

m_{1}Â = (- 1 â€“ 2)/(4 â€“ 1)

m_{1}Â = -1

Let slope of line AD be m_{2}

AD is perpendicular to BC

m_{1}Â Ã— m_{2}Â = -1

-1 Ã— m_{2}Â = -1

m_{2}Â = 1

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

y â€“ 3 = 1 Ã— (x â€“ 2)

y â€“ 3 = x â€“ 2

y â€“ x = 1

Equation of the altitude from vertex A = y â€“ x = 1

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

Equation of BC is

y + 1 = -1 Ã— (x â€“ 4)

y + 1 = -x + 4

x + y â€“ 3 = 0 â€¦â€¦â€¦â€¦â€¦â€¦â€¦(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 line i.e., Ax + By + C = 0, we get

Length of AD =Â

âˆ´ The equation and the length of the altitude from vertex A are y â€“ x = 1 and

âˆš2 units respectively.

**18. 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}Â **

**Solution:**

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

bx + ay = ab

bx + ay â€“ ab = 0 â€¦â€¦â€¦â€¦â€¦â€¦..(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 the sides we get

âˆ´ 1/p^{2} = 1/a^{2} + 1/b^{2}

Hence proved.