NCERT Solutions: Exercise 9.3- Straight Lines

# Exercise 9.3- Straight Lines NCERT Solutions | Mathematics (Maths) Class 11 - Commerce PDF Download

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

ANSWER : - (i) The given equation is x + 7y = 0.

It can be written as

This equation is of the form y = mx + c, where  .

Therefore, equation (1) is in the slope-intercept form, where the slope and the y-intercept are   and 0 respectively.

(ii) The given equation is 6x + 3y – 5 = 0.

It can be written as

Therefore, equation (2) is in the slope-intercept form, where the slope and the y-intercept are –2 and   respectively.

(iii) The given equation is y = 0.

It can be written as

y = 0.x + 0 … (3)

This equation is of the form y = mx + c, where m = 0 and c = 0.

Therefore, equation (3) is in the slope-intercept form, where the slope and the y-intercept are 0 and 0 respectively.

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.

ANSWER : - (i) The given equation is 3x + 2y – 12 = 0.

It can be written as

This equation is of the form  , where a = 4 and b = 6.

Therefore, equation (1) is in the intercept form, where the intercepts on the x and y axes are 4 and 6 respectively.

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

It can be written as

This equation is of the form  , where a =    and b = –2.

Therefore, equation (2) is in the intercept form, where the intercepts on the x and y axes are    and –2 respectively.

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

It can be written as

This equation is of the form  , where a = 0 and b =   .

Therefore, equation (3) is in the intercept form, where the intercept on the y-axis is   and it has no intercept on the x-axis.

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

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

⇒ 12x + 72 = 5y – 10

⇒12x – 5y + 82 = 0 … (1)

On comparing equation (1) with general equation of line Ax  + By + C = 0, we obtain A = 12, B = –5, and C = 82.

It is known that the perpendicular distance (d) of a line Ax  + By  + C = 0 from a point (x1, y1) is given by  .

The given point is (x1, y1) = (–1, 1).

Therefore, the distance of point (–1, 1) from the given line

Question 4: Find the points on the x-axis, whose distances from the line    are 4 units.

ANSWER : - The given equation of line is

On comparing equation (1) with general equation of line Ax  + By +  C = 0, we obtain 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.

It is known that the perpendicular distance (d) of a line Ax  + By  + C = 0 from a point (x1, y1) is given by  .

Therefore,

Thus, the required points on the x-axis are (–2, 0) and (8, 0).

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

ANSWER : - It is known that the distance (d) between parallel lines Ax + By + C1 = 0 and Ax + By + C2 = 0 is given by  .

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

Here, A = 15, B = 8, C1 = –34, and C2 = 31.

Therefore, the distance between the parallel lines is

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

lx +  ly +  p = 0 and lx  + lyr = 0

Here, A = l, B = l, C1 = p, and C2 = –r.

Therefore, the distance between the parallel lines is

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

ANSWER : - The equation of the given line is

, which is of the form y = mx + c

∴ Slope of the given line

It is known that parallel lines have the same slope.

∴ Slope of the other line =

Now, the equation of the line that has a slope of    and passes through the point (–2, 3) is

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

ANSWER : - The given equation of line is  .

, which is of the form y = mx + c

∴Slope of the given line

The slope of the line perpendicular to the line having a slope of    is

The equation of the line with slope –7 and x-intercept 3 is given by

y = m (xd)

y = –7 (x – 3)

y = –7x  21

⇒ 7x + y = 21

Question 8: Find angles between the lines

ANSWER : - The given lines are  .

The slope of line (1) is  , while the slope of line (2) is  .

The acute angle i.e., θ between the two lines is given by

Thus, the angle between the given lines is either 30° or 180° – 30° = 150°.

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.

ANSWER : - The slope of the line passing through points (h, 3) and (4, 1) is

The slope of line 7x – 9y – 19 = 0 or    is  .

It is given that the two lines are perpendicular.

Thus, the value of h is  .

Question 10: Prove that the line through the point (x1, y1) and parallel to the line Ax + By + C = 0 is A (x –x1)  + B (y – y1) = 0.

ANSWER : -  The slope of line A+ B C = 0 or    is

It is known that parallel lines have the same slope.

∴ Slope of the other line =

The equation of the line passing through point (x1, y1) and having a slope   is

Hence, the line through point (x1, y1) and parallel to line A+ B+ C = 0 is

A (x –x1)  + B (y – y1) = 0

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.

ANSWER : -  It is given that the slope of the first line, m1 = 2.

Let the slope of the other line be m2.

The angle between the two lines is 60°.

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

In this case, the equation of the other line is  .

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

In this case, the equation of the other line is  .

Thus, the required equation of the other line is   or   .

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

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

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

Accordingly, mid-point of AB

Slope of AB

∴Slope of the line perpendicular to AB =

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

Thus, the required equation of the line is 2x + y = 5.

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

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

Slope of the line joining (–1, 3) and (a, b), m1

Slope of the line 3x – 4y – 16 = 0 or

Since these two lines are perpendicular, m1m2 = –1

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

∴3a – 4b = 16 … (2)

On solving equations (1) and (2), we obtain

Thus, the required coordinates of the foot of the perpendicular are  .

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.

ANSWER : -  The given equation of line is y = mx + c.

It is given that the perpendicular from the origin meets the given line at (–1, 2).

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

∴Slope of the line joining (0, 0) and (–1, 2)

The slope of the given line is m.

Since point (–1, 2) lies on the given line, it satisfies the equation y = mx +  c.

Thus, the respective values of m and c are  .

Question 15: If p and q are the lengths of perpendiculars from the origin to the lines x cos θy sin θ = k cos 2θ and x sec θ + y cosec θ = k, respectively, prove that p2 + 4q2 = k2

ANSWER : -  The equations of given lines are

x cos θy sinθ = k cos 2θ … (1)

x secθ + y cosec θ= k … (2)

The perpendicular distance (d) of a line Ax +  By + C = 0 from a point (x1, y1) is given by  .

On comparing equation (1) to the general equation of line i.e., Ax  + By +  C = 0, we obtain A = cosθ, B = –sinθ, and C = –k cos 2θ.

It is given that p is the length of the perpendicular from (0, 0) to line (1).

On comparing equation (2) to the general equation of line i.e., Ax + By + C = 0, we obtain A = secθ, B = cosecθ, and C = ­–k.

It is given that q is the length of the perpendicular from (0, 0) to line (2).

From (3) and (4), we have

Hence, we proved that p2 + 4q2 = k2.

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.

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

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

(y – 3) = 1(x – 2)

xy + 1 = 0

yx = 1

Therefore, equation of the altitude from vertex A = yx = 1.

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

The equation of BC is

The perpendicular distance (d) of a line Ax  + By  + C = 0 from a point (x1, y1) is given by  .

On comparing equation (1) to the general equation of line Ax + By + C = 0, we obtain A = 1, B = 1, and C = –3.

Thus, the equation and the length of the altitude from vertex A are yx = 1 and   units respectively.

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  .

ANSWER : -  It is known that the equation of a line whose intercepts on the axes are a and b is

The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by  .

On comparing equation (1) to the general equation of line Ax + By + C = 0, we obtain A = b, B = a, and C = –ab.

Therefore, if p is the length of the perpendicular from point (x1, y1) = (0, 0) to line (1), we obtain

On squaring both sides, we obtain

Hence, we showed that  .

The document Exercise 9.3- Straight Lines NCERT Solutions | Mathematics (Maths) Class 11 - Commerce is a part of the Commerce Course Mathematics (Maths) Class 11.
All you need of Commerce at this link: Commerce

## Mathematics (Maths) Class 11

75 videos|238 docs|91 tests

## FAQs on Exercise 9.3- Straight Lines NCERT Solutions - Mathematics (Maths) Class 11 - Commerce

 1. What is the significance of exercise 9.3 in NCERT Solutions for Straight Lines JEE?
Ans. Exercise 9.3 in NCERT Solutions for Straight Lines JEE is significant because it focuses on solving problems related to straight lines, which is an important topic in JEE (Joint Entrance Examination). This exercise provides a comprehensive understanding of various concepts and techniques used in solving problems related to straight lines.
 2. How can I use NCERT Solutions Exercise 9.3 to prepare for JEE?
Ans. NCERT Solutions Exercise 9.3 can be used as a valuable resource for JEE preparation. By practicing the problems and understanding the solutions provided in this exercise, you can enhance your problem-solving skills and gain a deeper understanding of the concepts related to straight lines. It is recommended to solve the problems step-by-step, referring to the solutions provided, and identify any gaps in your understanding.
 3. Are the solutions in NCERT Exercise 9.3 for Straight Lines JEE accurate?
Ans. Yes, the solutions provided in NCERT Exercise 9.3 for Straight Lines JEE are accurate. These solutions are carefully crafted by subject matter experts who have extensive knowledge and experience in the field. However, it is always advisable to cross-verify the solutions and seek additional help if you encounter any doubts or difficulties in understanding the concepts or solutions.
 4. Can I solely rely on NCERT Exercise 9.3 for Straight Lines JEE preparation?
Ans. While NCERT Exercise 9.3 for Straight Lines JEE is a valuable resource, it is recommended to supplement your preparation with additional study materials, reference books, and online resources. The JEE syllabus is vast, and relying solely on one resource may not provide you with a comprehensive understanding of all the topics. It is important to explore different sources and practice a variety of problems to strengthen your knowledge and problem-solving skills.
 5. How can I make the most of NCERT Exercise 9.3 for Straight Lines JEE?
Ans. To make the most of NCERT Exercise 9.3 for Straight Lines JEE, follow these tips: 1. Read the theory and concepts related to straight lines from the NCERT textbook before attempting the exercise. 2. Solve each problem step-by-step, referring to the solutions provided, and understand the logic and techniques used. 3. Identify any gaps in your understanding and revisit the corresponding theory or concepts. 4. Practice additional problems from other books or online resources to strengthen your problem-solving skills. 5. Seek help from teachers or online forums if you encounter any difficulties or have doubts in understanding the solutions.

## Mathematics (Maths) Class 11

75 videos|238 docs|91 tests

### Up next

 Explore Courses for Commerce exam

### Top Courses for Commerce

Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;