NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines (Ex 10.3) Exercise 10.3

The Central Board of Secondary Education is one of the most highly reputed and esteemed educational boards that operate in India. There are a multitude of schools spread all over the country that are affiliated with the board. CBSE follows an extremely objective approach to its curriculum. CBSE wants its students to have a holistic understanding of a subject. All the concepts that are discussed in the CBSE syllabus must be thoroughly understood by the student for them to be able to answer the questions that are asked in the exams. For students to score well in the CBSE exams, they must have a very clear and solid understanding of the subject and the ideas that they are discussing. CBSE is widely known for its vast syllabus, and students often fear the size of the syllabus, but experts at Extramarks reassure students that there is nothing for them to feel overwhelmed about. Although the transition from a Class 10 level to a 10+2 level is challenging for many students. The shift in the scale of difficulty during the transition is very distinct, although experts at Extramarks have shared really easy ways through which students can easily adjust. CBSE prioritises ideas and concepts and their applications over memorising and learning. The curriculum of the CBSE is very application based and therefore, scoring well in the exams is contingent on a very in-depth understanding of the ideas discussed in the chapter. Top educators at Extramarks want all students to understand that the new academic atmosphere might be daunting, but there are effective ways that can be employed to deal with it.

The National Council of Educational Research and Training (NCERT) is a central organisation that was established in 1961 by the Indian government. NCERT legislates and controls all the issues related to the education of the students. NCERT assists the Central Government and the State Governments on issues that are related to the educational development and the academic dexterity of the country. NCERT has set up its guidelines andcurated its syllabus. NCERT and CBSE work together symbiotically. NCERT asks all the CBSE schools to follow their syllabus and guidelines. NCERT also has books published that are based on its syllabus and guidelines. NCERT recently have been working towards expanding its reach to improve the educational environment of the country. NCERT has introduced many amenities for students, like foreign exchange programmes, journals in which students can get themselves published, and various others. NCERT trains the students in the best way possible, and therefore they follow a rigorous academic profile. Students often find this very daunting but experts at Extramarks always reassure the students by providing them with really useful study materials and hacks. All educators at Extramarks have reiterated the importance of a consistent practice of problems in a particular chapter, steadily moving ahead with the syllabus, and regularly revising all the problems that a student has done before. Teachers have observed that once a student follows these hacks, they see significant improvement in them.

Mathematics is one of the core subjects of any student from Class 11 who has chosen Science or Commerce as their stream of study. Science has been considered a difficult field primarily because it is highly practical, analytical, and intuitive. A student can excel in science with a diligent attitude toward their studies and patience. The ideas that are included in the Class 11 Mathematics curriculum have various applications. Not only do students have to prepare for mathematics, but at a higher secondary level, the concepts that are taught in Mathematics echo in other science subjects like Physics and Chemistry. Concepts from Mathematics echo widely in Physics to an extent that almost half of the Physics question papers are numerical-based questions. It is observed that students must take Mathematics very seriously. A student is given their final exam grade, which is calculated by taking the average of their grades in other subjects. Therefore, students have to be consistent about their performance throughout all the subjects. A student must have all the fundamentals clear. The question paper compiled by the CBSE has analytical questions based on the applications of all the concepts that a student learns.

Chapter 10 in the CBSE curriculum is Straight Lines. Straight Line is a part of the bigger and extremely important section called Coordinate Geometry. Coordinate geometry is one of the sections of the Maths Syllabus that has implications for both Chemistry and Physics therefore, educators at Extramarks consider this chapter to be one of the most important chapters in the Syllabus. Straight lines are the introductory chapter to concepts that will give rise to concepts like Parabola, Hyperbola, Circle, Ellipse, etc. Teachers ask the students to lay a lot of emphasis on this chapter. Therefore, Extramarks has released the NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3. The NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3 are keys and answers to all the questions in the NCERT Textbook.

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NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines (Ex 10.3) Exercise 10.3

A considerable number of students shift from other boards to CBSE. The CBSE Board is highly popular among students who are preparing for competitive exams like JEE Advanced. JEE Mains, NEET, etc. The reason behind this is that the direct and objective study technique completely overlaps with the multiple-choice questions that are asked in these competitive exams. Therefore, when students are referring to NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3 they are not only getting prepared for their CBSE exams but also for the competitive exams as well. A significant amount of questions come from these sections and therefore students must have all the basic and elemental ideas of the chapter very clear. The NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3, help students get accustomed to the kind of questions that generally come in the exam. The NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3 also helps the student understand the right and acceptable way of writing answers for CBSE. The competitive exams are extremely difficult, and the NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3, helps students build a very solid foundation in the subject. Top educators at Extramarks have repeated the importance of a calm mind and consistent behaviour as the most important qualities that a student must learn to inculcate in their studying habits.

Access NCERT Solutions for Class 11 Mathematics Chapter 10-Straight Lines

The NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3, have been a great resource for students. CBSE has a provision where students can appear in their exams in either Hindi or English. Hindi is the national language of India, and the demographic of people who speaks Hindi is huge. Therefore Extramarks also releases NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3 in Hindi. There is a general scarcity of credible and reliable resources in Hindi, and therefore NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3 is an extremely important tool a student studying in a Hindi-medium school can own. Students who refer to the NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3, in Hindi get accustomed to the distinct mathematical vocabulary in that language. Students who are appearing in their CBSE exams in Hindi when referring to the NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3 in Hindi are introduced to a lot of additional information. These students get an idea of the correct way in which students must answer the questions. NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3 is provided by highly qualified professionals at Extramarks who are well-versed and fluent in speaking Hindi. These teachers also have years of experience in teaching students Hindi. Therefore, it is again reassured that students can rely on the solutions blindly.

NCERT Solutions for Class 11 Maths Chapters

Experts at Extramarks have worked tirelessly to provide the most appropriate and suitable answers for NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3. Extramarks releases the NCERT Solutions as a way to assist students in their time of need. Students in Class 11 of the CBSE greatly benefit from the solutions. The solutions are curated keeping in mind the various kinds of students who would access them. NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3 are written and compiled in a specific way to make sure that students from varied academic backgrounds would be able to use it. The explanations provided by Extramarks are lucidly written, ensuring that everyone can comprehend them. There is a Hindi version of the NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3, which provides assistance to students who would be attempting the exam in Hindi. Students often find the NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3 sufficient, but sometimes doubts persist. Students can access Extramarks’ website or the mobile application for more resources, like the NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3. Experts at Extramarks have compiled other resources like live classes, recorded classes, test series, quizzes, options for live chat, and detailed performance reports and analysis. Extramarks provides holistic support to as many students as possible.

Students can access the NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3 from the Extramarks’ website.

NCERT Solution Class 11 Maths of Chapter 10 All Exercises

NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3, are the solutions to the third exercise of Chapter 10. There are a total of four exercises in the chapter. There are a total of 18 questions in the third exercise, and the solutions to all 18 questions are in the NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3. The NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3, are compiled in a way that makes sure that all the questions have extremely simple step-by-step explanations. The NCERT textbook has the sums arranged in increasing order of difficulty, and therefore the explanations have to be followed in chronology to get a smooth introduction to the complicated problems. NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3 follows the chronology in which they are given in the NCERT textbook. The question number as well as the final answer are tallied with those in the textbook.

Topics Covered in Exercise 10.3 NCERT

Chapter 10 of the CBSE Class 11 curriculum of Mathematics is Straight lines. Students learn about the equation of a straight line as represented by Ax+By+C = 0. The capital letters represent the numerical constant of the equation. They also learn about the different kinds of straight lines and their different forms. Students learn to calculate the distance from one point to a straight line. The last thing discussed in the chapter is the distance between two parallel lines.

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Exercise 10.3

NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3 make sure that the solutions are correct and the explanations are correct. The NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3 strictly follows the CBSE guidelines. Extramarks provide multiple other resources to endure the optimum performance from the student’s end. Understanding complicated concepts depend on students getting the correct form of guidance. Students get a chance to have a thorough understanding of these concepts. Top educators at Extramarks having years of relevant experience have considered the NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3 to be an extremely trustworthy and reliable resource.

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Q.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.

Ans.

(i)The given equation​ is:      x+7y=0⇒y=−17x+0Which is in the form of y=mx+c,where m=−17and C=0.(ii)The given equation​ is: 6x+3y−5=0⇒        3y=−6x+5⇒          y=−2x+53Which is in the form of y=mx+c, where m=−2 and C=53.(iii)The given equation​ is:          y=0⇒         y=0.x+0Which is in the form of y=mx+c, where m=0 and C=0.

Q.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.

Ans.

(i)The given equation is: 3x+2y−12=0 ⇒     3x+2y=12Dividing both sides by 12, we get        3x12+2y12=1⇒      x4+y6=1Which is in the intercept form and x,y- intercepts are 4 and 6 respectively.(ii)The given equation is:        4x−3y=6 Dividing both sides by 6, we get       4x6−3y6=66 ⇒  x(32)−y2=1 ⇒ x(32)+y−2=1Which is in the intercept form and x,y- intercepts are 32 and−2 respectively.(iii)The given equation is:                 3y+2=0

⇒       y=−23⇒  y(−23)=1Here, y-intercept is −23 and no intercept with x-axis.

Q.3

Reduce the following equations into normal form. Find their perpendicular distances from the originand angle between perpendicular and the positive x-axis.(i) x – 3 y + 8 = 0,(ii) y – 2 = 0,(iii) x–y = 4.

Ans.

(i) x−3 y+8=0        −x+3 y=8Dividing both sides by 12+(−3)2=2, we get          −12x+32 y=82⇒xcos2π3+ysin2π3=4Which is in the normal form i.e., xcosω+ysinω=p,where perpendicular distance(p)=4and angle between perpendicular         and x-axis(ω)=2π3

(ii) y−2=0⇒   0.x+ y=2Dividing both sides by 02+(1)2=1, we get⇒      01x+ y1=2⇒x cos 90°+y sin 90° =2Which is in the normal form i.e., x cosω+y sinω=p,where perpendicular distance(p)=2and angle between perpendicular          and x-axis(ω)=90°(iii) x−y=4Dividing both sides by 12+(−1)2=2, we get⇒            12x− y2=42⇒             xcos45°−y sin45°=22⇒      xcos(360°−45°)+y sin(360°−45°)=22⇒              xcos(315°)+y sin(315°)=22Which is in the normal form i.e., xcos ω+y sin ω=p,where perpendicular distance(p)=22and angle between perpendicular          and x-axis (ω)=315°

Q.4 Find the distance of the point (–1, 1) from the line 12(x + 6) = 5(y – 2).

Ans.

The equation of given line is:

12 (x+6)=5(y−2)⇒   12x+72=5y−10⇒  12x−5y+ 82=0Distance of the line from the point (−1,1),        d=|12(−1)−5(1)+ 82(12)2+(−5)2|⇒       d=|−12−5+ 82144+25|   =6513⇒       d=5 unitsThus, the distance of given line from the point (−1,1) is 5 units.

Q.5

Find the points on the x-axis, whose distances from the line x3+y4=1  are 4 units.

Ans.

Let the point on x-axis be (a, 0).Distance of the line x3+y4=1 from the point (a,0), d=|a3+04−1(13)2+(14)2|

⇒ 4=|a3+04−1(19)+(116)|⇒ ± 4=a3−1(16+99×16)⇒  ± 4(512)=a3−1⇒    ± (53)+1=a3Taking  (+) sign,   5+33=a3⇒ a=8So, the point on x-axis is (8,0).Taking  (−) sign,      −5+33=a3⇒    a=−2So, the point on x-axis is (−2,0).Thus, the distance of line from point (−2,0) and (8,0) is 4.

Q.6 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.

Ans.

(i) Given parallel lines are: 15x+8y−34=0 and 15x+8y+31=0Distance between two parallel lines, d=|C1−C2A2+B2|where, A=15, B=8, C1=−34 and C2=31∴   d=|−34−31152+82|     =|−65289|     =6517 unitsThus, the distance between two given parallel lines is 6517 units.(ii)Given parallel lines are:l(x+y)+p = 0 and l(x+y)−r = 0Distance between two parallel lines, d=|C1−C2A2+B2|where, A=l, B=l, C1=p and C2=− r∴   d=|p+rl2+l2|

=|p+r2l2| =12|p+rl| unitsThus, the distance between two given parallellines is 12|p+rl| units.

Q.7 Find equation of the line parallel to the line 3x − 4y + 2 = 0 and passing through the point (–2, 3).

Ans.

Let the equation of the line parallel to the line 3x−4y+2=0 is3x−4y+λ=0 and it is passing through the point (−2,3),then 3(−2)−4(3)+λ=0⇒          − 6−12+λ=0⇒                 −18+λ=0⇒       λ=18Thus, parallel line is 3x−4y+18=0.

Q.8 Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having x intercept 3.

Ans.

The given equation of line  is x−7y+5=0.Then, given equation of line can be written as:y=17x+57, comparing with y=mx+c, we havem=17

Slope of perpendicular line, M=−1m=−7Equation of perpendicular line is: y=Mx+C’ y=−7x+C‘Since, perpendicular has x-intercept 3. So,perpendicular passes through the point (3,0).  0=−7(3)+C‘⇒      C‘=21Thus, the equation of perpendicular line is:   y=−7x+21​ or 7x+y−21=0

Q.9

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

Ans.

Thegiven linesare 3 x+y=1⇒y=− 3 x+1 ...(i)and   x+3y=1⇒y=−13x+13  ...(ii) The slope of equation (i),m1=−3The slope of equation (ii),m2=−13Angle between two lines is θ.Then,tanθ=|m2−m11+m1m2|  

=|−13−(−3)1+(−3)(−13)| =|−1+331+1| =|232|tanθ=±13tanθ=13 or −13⇒tanθ=tan30° or tan150°⇒       θ=30° or 150°Thus, the angle between two lines is 30° or 150°.

Q.10 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.

Ans.

Slope of the line through the points (h,3) and (4, 1),m1=1−34−hSlope of the line 7x−9y+19=0,      m2=−7−9=79

Since, both lines are perpendicular to each other, som1×m2=−1        1−34−h×79=−1⇒        −14=−(36−9h)⇒            14=36−9h⇒               h=36−149 =229Thus, the value of h is 229.

Q.11 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.

Ans.

Slope of the line Ax+By+C=0, m=−ABEquation of parallel line to given line is:y=−ABx+C‘  ...(i)Equation (i) is passing through the point (x1,y1), then from equation (i) y1=−ABx1+C‘⇒C‘=ABx1+y1

Putting value of C’ in equation (i), we get y=−ABx+ABx1+y1⇒By=−Ax+Ax1+By1   A(x−x1)+B(y−y1)=0Which is the required equation of parallel line.

Q.12 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.

Ans.

Let slope of one line, m1=2 and slope of other line, m2=m.Angle between two lines=60°Since, tanθ=|m2−m11+m1m2|⇒  tan 60°=|m−21+2m|⇒          3=|m−21+2m|⇒      ±3=m−21+2m

For (+)ve sign: 3=m−21+2m        3(1+2m)=m−2        3+23m=m−2        (23−1)m=−(2+3)              m=−(2+3)(23−1)Equation of line passing through the point (2,3):              y−3=−(2+3)(23−1)(x−2)     (y−3)(23−1)=−(2+3)x+2(2+3)(2+3)x+(23−1)y=2(2+3)+3(23−1)(2+3)x+(23−1)y=4+23+63−3(3+2)x+(23−1)y=1+83For (−) ve sign:       −3=m−21+2m −3(1+2m)=m−2−3−23 m=m−2   2−3=(23+1)m                          

m=2−3(23+1)Equation of line passing through the point (2,3):                   y−3=2−3(23+1)(x−2)⇒(23+1)y−3(23+1)=(2−3)x−4+23⇒    (3−2)x+(23+1)y=63+3−4+23⇒    (3−2)x+(23+1)y=83−1Therefore, the equation of other line is (3+2)x+(23−1)y=1+83 or  (3−2)x+(23+1)y=83−1.

Q.13 Find the equation of the right bisector of the line segment joining the points (3, 4) and (–1, 2).

Ans.

Mid−point of (3,4) and (−1,2)=(3−12,4+22)         =(1,3)Slope of line joining the points (3,4) and (−1,2)         =2−4−1−3         =−2−4         =12

Slope of right bisector of line=−1(12)=−2Equation of line passing throgh the point (1,3) and having slope −2. y−3=−2(x−1) y−3=−2x+22x+y=5Thus, the equation of right bisector is 2x+y=5.

Q.14 Find the coordinates of the foot of perpendicular from the point (–1, 3) to the line 3x – 4y – 16 = 0.

Ans.

Given line is:  3x−4y−16=0⇒ 3x−4y=16 ...(i)Slope of the line 3x−4y−16=0,m=−3−4[∵m=−AB]  m=34Slope of perpendicular to the line=−1(34)     =−43

Equation of perpendicular through the point (−1,3) is     y−3=−43(x+1)⇒   3y−9=−4x−4⇒         4x+3y=9−4⇒         4x+3y=5 ...(ii)Solving equation (i) and equation (ii), we getx=6825 and y=−4925Thus, the coordinates of the foot of perpendicular is (6825,−4925).

Q.15 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.

Ans. 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.

Q.16 If p and q are the lengths of perpendiculars from the origin to the lines x cosθ − y sin θ = k cos 2θ and x sec θ + y cosec θ = k, respectively, prove that p2 + 4q2 = k2.

Ans.

The given lines are:   xcosθ−ysin θ=kcos 2θ ...(i)and  xsecθ+ycosecθ=k          ...(ii)p=Length of perpendicular from origin to line (i)    =|(0)cosθ−(0)sin θ−kcos 2θcos2θ+sin2θ|p=kcos 2θq=Length of perpendicular from origin to line (ii)    =|(0)secθ+(0)cosecθ−ksec2θ+cosec2θ|

    =kcos2θ+sin2θcos2θ×sin2θq=ksinθ cosθcos2θ+sin2θ    =ksinθcosθ  L.H.S.=p2+4q2    =(kcos 2θ)2+4(ksinθ cosθ)2    =k2(cos 2θ)2+k2(2sinθ cosθ)2    =k2(cos22θ+sin22θ)    =k2×1    =k2=R.H.S.

Q.17 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.

Ans.

Slope of BC=2+11−4      =−33      =−1Slope of perpendicular on BC=−1−1      =1Equation of altitude on BC through A(2,3)     y−3=1(x−2)

⇒    y−3=x−2⇒    x−y=−3+2⇒    y−x=1Equation of line through BC     y+1=2+11−4(x−4)     y+1=3−3(x−4)     y+1=−1(x−4)⇒   x+y=3Length of perpendicular from A(2,3) to BCd=|2+3−312+(−1)2|    =22 d=2Thus, equation of perpendicular is y−x=1 and length of perpendicular is 2.

Q.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  1p2=1a2+1b2.

Ans.

Equation of line having intercepts a and b with axis, xa+yb=1 ...(i) 

      p=Length​ of perpendicular from origin to line (i)    =|0a+0b−1(1a)2+(1b)2|            p=|11a2+1b2|Squarring both sides, we get           p2=(11a2+1b2)2         p2=1(1a2+1b2)⇒1a2+1b2=1p2Therefore, 1p2=1a2+1b2  has proved.

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FAQs (Frequently Asked Questions)

1. Where can students find the NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3?

The NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3 are available on the Extramarks’ website and they can also be accessed through their mobile application as well. The online platform allows the students to access the solutions and other resources that Extramarks provide anytime and anywhere. This is of great help to the students because they are constantly under the guidance of top educators. Maths Class 11 Chapter 10 Exercise 10.3 can be accessed through Extramarks’ website.

2. How can a student get help with NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3?

NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3 has been considered to be worthy of assistance by teachers and students alike. Students can use these solutions anytime during their studying. Students can refer to them while solving the exercise for the first time. Students can also use them when they are revising and doing the sums again. Although it is noticed that students use the solutions the most just before the exams. The solutions help students figure out their mistakes and therefore it helps them in becoming confident and self-reliant. This helps students keep calm during the exam which reduces the chances of making mistakes and silly errors. Class 11th Maths Exercise 10.3 is a tricky exercise and the teachers at Extramarks have done flawless work for these solutions.

3. How many exercises do Extramarks provide the solutions for?

NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3 is the solution of the third exercise. There are a total of four exercises in the entire chapter the last one being a miscellaneous exercise. Extramarks has solutions for all the exercises. Each solution goes through the same process of scrutiny and review making sure the explanations are eloquent and comprehensible. There are even solutions to the NCERT textbook provided in Hindi. Extramarks has provided solutions for every subject in the CBSE curriculum for every class. Ex 10.3 Class 11 is a difficult exercise but the solutions are easy and direct making sure that they can be used by everyone.