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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Ten Exemplar Solution

Find the equation of the line which passes through the point (–4, 3), and the portion of the line intercepted between the axes is divided internally in the ratio 5: 3 by this point.

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Vishal Baghel

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t A B b e a l i n e t h r o u g h a (- 4, 3) a n d m e e t s x a x i s a t A (a, 0) a n d \\ y a x i s a t B (0, b) . \\ ∴ - 4 = \frac{5 \times 0 + 3 a}{5 + 3} \\ \Rightarrow - 4 = \frac{3 a}{8} \Rightarrow a = \frac{- 32}{3} [\begin{array}{l} ? X = \frac{m_{1} x_{2} + m_{2} x_{1}}{m_{1} + m_{2}} \\ a n d Y = \frac{m_{1} y_{2} + m_{2} y_{1}}{m_{1} + m_{2}} \end{array}] \\ a n d 3 = \frac{5 \times b + 3 \times 0}{5 + 3} \\ \Rightarrow 3 = \frac{5 b}{8} \Rightarrow b = \frac{24}{5} \\ I n t e r c e p t f o r m o f t h e l i n e i s \\ \frac{x}{\frac{- 32}{3}} + \frac{y}{\frac{24}{5}} = 1 \\ \Rightarrow \frac{- 3 x}{32} + \frac{5 y}{24} = 1 \\ \Rightarrow - 9 x + 20 y = 96 \Rightarrow 9 x - 20 y + 96 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s 9 x - 20 y + 96 = 0 . \end{array}$

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9 months ago

Mangalayatan University Colleges in Uttar Pradesh

How can I apply for Mangalayatan University if I have Cleared CUET 2025 exam?

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Saurabh Khanduri

Contributor-Level 10

If you are a candidate who has cleared the CUET exam 2025, you can visit the official website of Mangalayatan University and fill out the registration form. You will get the option to mention your CUET score which will allow you to proceed to the personal interview round.

You will also be required to attach your CUET scorecard and bring the original copy of the same for document verification.

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9 months ago

Tver State Medical University Study Abroad

What is the average salary of Tver State Medical University graduates?

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Nishtha Sharma

Contributor-Level 10

Tver State Medical University is one of the most esteemed universities in Russia for pursuing higher education. As per unofficial sources, the average salary of Tver State Medical University graduate is RUB 720,000 (INR 7.76 L), annually. The university graduates are highly skilled in Medical Education, Teaching, Communication, Medicine, Healthcare, etc. According to sources available on the web, the Tver State Medical University placement rate is over 90%, and the graduates are employed in hospital and healthcare institutions.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Ten Exemplar Solution

A straight line moves so that the sum of the reciprocals of its intercepts made on axes is constant. Show that the line passes through a fixed point.

(Hint: 1/a+1/b=1, where a + b =” constant” =1/k (say). This implies that a + b = k. Line passes through the fixed point (k, k).)

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Vishal Baghel

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} I n t e r c e p t s f o r m o f a s t r a i g h t l i n e i s \\ \frac{x}{a} + \frac{y}{b} = 1 \\ w h e r e a a n d b a r e t h e m a d e b y t h e l i n e o n t h e a x e s . \\ G i v e n t h a t : \frac{1}{a} + \frac{1}{b} = \frac{1}{k} (s a y) \\ \Rightarrow \frac{k}{a} + \frac{k}{b} = 1 \\ w h i c h s h o w s t h a t t h e l i n e i s t h r o u g h t h e f i x e d (k, k) . \end{array}$

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9 months ago

Tver State Medical University Study Abroad

What skills Tver State Medical University graduates prioritizes for placement?

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Ishita Kalra

Contributor-Level 10

Tver State Medical University is one of top universities for higher education in Russia. The graduates are employed in hospital and healthcare institutions, as per unofficial sources. Tver University graduates are constantly in high demand all over the world. Tver State Medical University graduates are highly skilled in the following:

Medical Education
Teaching
Communication
Medicine
Healthcare
Medical Devices
Interpersonal Skills
Teamwork
Medical Diagnostics
IELTS
Clinical Research
Emergency Management
Public Health
Mediation
Healthcare Industry

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9 months ago

Study Abroad India

Can I carry my mobile phone or calculator to the Jharkhand Paramedical entrance exam?

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Nishtha Singh

Beginner-Level 5

Absolutely not. Items such as mobile phones, calculators, smart watches, pagers, electronic dictionaries, or log tables are strictly prohibited. Invigilators may confiscate them upon entry for the Jharkhand Paramedical entrance exam. Applicants are advised to visit the exam centre with stationary item such as pen only.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Ten Exemplar Solution

In what direction should a line be drawn through the point (1, 2) so that its point of intersection with the line x + y = 4 is at a distance √5 from the given point?

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Vishal Baghel

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t t h e g i v e n l i n e x + y = 4 a n d t h e r e q u i r e d l i n e' l' a t B (a, b) \\ S l o p e o f l i n e' l' i s g i v e n b y m = \frac{b - 2}{a - 1} = t a n θ \dots (i) \\ G i v e n t h a t A B = \frac{\sqrt[]{6}}{3} \\ S o, b y f o r m u l a f o r A (1, 2) a n d B (a, b) \\ \sqrt[]{{(a - 1)}^{2} + {(b - 2)}^{2}} = \frac{\sqrt[]{6}}{3} \\ O n s q u a r i n g b o t h t h e s i d e \\ a^{2} + 1 - 2 a + b^{2} + 4 - 4 b = \frac{6}{9} \\ \Rightarrow a^{2} + b^{2} - 2 a - 4 b + 5 = \frac{6}{9} \dots (i i) \\ B (a, b) a l s o s a t i f i e s t h e e q n . x + y = 4 \\ ∴ a + b = 4 \dots (i i i) \\ O n s o l v i n g (i i) a n d (i i i), w e g e t a = \frac{3 \sqrt[]{3} + 1}{2 \sqrt[]{3}}, b = \frac{5 \sqrt[]{3} - 1}{2 \sqrt[]{3}} \\ P u t t i n g t h e v a l u e s o f a a n d b i n e q n . (i), w e h a v e \\ t a n θ = \frac{\frac{5 \sqrt[]{3} - 1}{2 \sqrt[]{3}} - 2}{\frac{3 \sqrt[]{3} + 1}{2 \sqrt[]{3}}} = \frac{5 \sqrt[]{3} - 1 - 4 \sqrt[]{3}}{3 \sqrt[]{3} + 1 - 2 \sqrt[]{3}} = \frac{\sqrt[]{3} - 1}{\sqrt[]{3} + 1} \\ ∴ t a n θ = t a n 1 5^{0} \Rightarrow θ = 1 5^{0} \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Ten Exemplar Solution

A variable line passes through a fixed-point P. The algebraic sum of the perpendiculars drawn from the points (2, 0), (0, 2), and (1, 1) on the line is zero. Find the coordinates of the point P.

(Hint: Let the slope of the line be m. Then the equation of the line passing through the fixed-point P (x?, y?) is y – y ? = m (x – x?). Taking the algebraic sum of perpendicular distances equal to zero, we get y – 1 = m (x – 1). Thus (x?, y?) is (1, 1).

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Vishal Baghel

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t (x_{1}, y_{1}) b e t h e c o o r d i n a t e s o f t h e g i v e n P a n d m b e t h e s l o p e o f t h e l i n e . \\ ∴ E q u a t i o n o f t h e l i n e i s y - y_{1} = m (x - x_{1}) \dots (i) \\ G i v e n a r e A (2, 0), B (0, 2) a n d C (1, 1) . \\ P e r p e n d i c u l a r f r o m A (2, 0) d_{1} (s a y) \\ d_{1} = \frac{0 - y_{1} - m (2 - x_{1})}{\sqrt[]{1 + m^{2}}} \\ P e r p e n d i c u l a r f r o m B (0, 2) d_{2} (s a y) \\ d_{2} = \frac{2 - y_{1} - m (0 - x_{1})}{\sqrt[]{1 + m^{2}}} \\ S i m i l a r l y, p e r p e n d i c u l a r f r o m C (1, 1) d_{3} (s a y) \\ d_{3} = \frac{1 - y_{1} - m (1 - x_{1})}{\sqrt[]{1 + m^{2}}} \\ A c c o r d i n g t o t h e q u e s t i o n, w e h a v e \\ d_{1} + d_{2} + d_{3} = 0 \\ ∴ \frac{0 - y_{1} - m (2 - x_{1})}{\sqrt[]{1 + m^{2}}} + \frac{2 - y_{1} - m (0 - x_{1})}{\sqrt[]{1 + m^{2}}} + \frac{1 - y_{1} - m (1 - x_{1})}{\sqrt[]{1 + m^{2}}} = 0 \\ \Rightarrow - y_{1} - 2 m + m x_{1} + 2 - y_{1} + m x_{1} + 1 - y_{1} - m + m x_{1} = 0 \\ \Rightarrow 3 m x_{1} - 3 y_{1} - 3 m + 3 = 0 \Rightarrow m x_{1} - y_{1} - m + 1 = 0 \\ t h e (1, 1) s a t i f i e s t h e a b o v e e q u a t i o n . \\ H e n c e, t h e (1, 1) l i e s o n t h e l i n e . \end{array}$

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9 months ago

IIT Bhubaneswar - Indian Institute of Technology

Can I get admission to IIT Bhubaneswar with a rank of 10,000?

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Sanjana Singh

Contributor-Level 6

Yes, it is possible to get admission to IIT Bhubaneswar with a rank of 10,000 in JEE Advanced. The closing ranks as per the IIT Bhubaneswar JEE Advanced cutoff 2025 varied between 3855 and 14467, for the General AI category candidates. A few specialisations in which candidates can easily get admission at 10,000 rank are B.Tech. in Engineering Physics and B.Tech. in Metallurgical and Materials Engineering. Some of the most competitive specialisation among the General AI category turned out to be BTech in Computer Science and Engineering and B.Tech. in Mathematics and Computing, for which it would not be possible to get admissi

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9 months ago

Engineering B.E. Eligibility

What is the academic requirement to be eligible for SUCET?

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Sumridhi Dhyani

Contributor-Level 7

For SUCET, candidates should have completed 12th class exam from a recognised board. In qualifying exam, candidates should have scored minimum 60% in Physics, Chemistry and Mathematics. Candidates who have done open schooling after 10th class will also be eligible for SUCET.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Ten Exemplar Solution

If the equation of the base of an equilateral triangle is x + y = 2 and the vertex is (2, –1), then find the length of the side of the triangle.

(Hint: Find the length of the perpendicular (p) from (2, –1) to the line and use p = l sin 60°, where l is the length of the side of the triangle.)

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Vishal Baghel

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} E q u a t i o n o f t h e b a s e A B o f a Δ A B C i s x + y = 2 \\ I n Δ A B D, \\ s i n 6 0^{0} = \frac{A D}{A B} \Rightarrow \frac{\sqrt[]{3}}{2} = \frac{A D}{A B} \Rightarrow A D = \frac{\sqrt[]{3}}{2} A B \\ L e n g t h o f p e r p e n d i c u l a r f r o m A (2, - 1) t o t h e l i n e x + y = 2 i s \\ A D = | \frac{1 \times 2 + 1 \times - 1 - 2}{\sqrt[]{{(1)}^{2} + {(1)}^{2}}} | \\ \Rightarrow \frac{\sqrt[]{3}}{2} A B = | \frac{2 - 1 - 2}{\sqrt[]{2}} | = | \frac{- 1}{\sqrt[]{2}} | \\ \Rightarrow \frac{\sqrt[]{3}}{2} A B = \frac{1}{\sqrt[]{2}} \Rightarrow A B = \frac{\sqrt[]{2}}{\sqrt[]{3}} \\ H e n c e, t h e r e q u i r e d l e n g t h o f s i d e = \sqrt[]{\frac{2}{3}} . \end{array}$

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9 months ago

Exam MUJEE

Can I still apply for the MUJEE 2025 exam?

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Saurabh Khanduri

Contributor-Level 10

Yes, it is stll possible to apply for the MUJEE exam. The last date to apply for the MUJEE 2025 exam is within the first week of July. Candidates can apply for the MUJEE exam by filling out the registration form on the official website of Mangalayatan University.

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9 months ago

IIT Bhubaneswar - Indian Institute of Technology

How much rank should I score to get IIT Bhubaneswar?

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9 months ago

KAU - Kerala Agricultural University Business Management Studies

Does Kerala Agricultural University offers MBA courses?

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Atul Pruthi

Contributor-Level 9

Yes, Kerala Agricultural University does offer an MBA course for the duration of two years, and it is divided into four semesters. The MBA courses at KAU are offered through the Faculty of Agriculture. The university only offers one specialisation in the MBA course, i.e., Agri Business Management. The requirement for admission to these courses is that candidates must complete their graduation from a related/ unrelated field with a certain minimum percentage/ CGPA.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Nine Maths

Sol:

$\begin{array}{l} G i v e n t h a t \underset{h \to 0}{l i m} \frac{\sqrt[]{x + h} - \sqrt[]{x}}{h} \\ = \underset{h \to 0}{l i m} \frac{\sqrt[]{x + h} - \sqrt[]{x}}{h} \times \frac{\sqrt[]{x + h} + \sqrt[]{x}}{\sqrt[]{x + h} + \sqrt[]{x}} [R a t i o n a l i z i n g t h e dghdhg] \\ = \underset{h \to 0}{l i m} \frac{x + h - x}{h [\sqrt[]{x + h} + \sqrt[]{x}]} = \underset{h \to 0}{l i m} \frac{h}{h [\sqrt[]{x + h} + \sqrt[]{x}]} \\ = \underset{h \to 0}{l i m} \frac{1}{\sqrt[]{x + h} + \sqrt[]{x}} \\ T a k i n g, w e h a v e \\ \frac{1}{\sqrt[]{x} + \sqrt[]{x}} = \frac{1}{2 \sqrt[]{x}} . \end{array}$

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9 months ago

Somaiya Vidyavihar University Eligibility Somaiya Vidyavihar University

What is the eligibility criteria for admission at Somaiya Vidyavihar University?

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Rashmi Kumar

Contributor-Level 10

Somaiya Vidyavihar course admissions for all of its courses are done based on entrance exams. Students can check the list of course eligibility for various courses from the table presented below:

Courses	Eligibility
BDes	Class 12 with 50% aggregate
BE/BTech	Class 12 with 45% with any board with English, Physics, and Mathematics as compulsory subjects
MA	Graduation with 55% aggregate marks in undergraduate examination
BSc	Class 12 with 50-55% aggregate
BCA	Class 12 with 55% aggregate
BBA	Class 12 with 55% aggregate
BCom	Class 12 with 55% aggregate
BBM/BMS	Class 12 with 55% aggregate
ME/MTech	Graduation with at least 50% marks (at least 45% marks in case of persons with disability)

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9 months ago

SUCET Application Form SUCET

What is the age limit for SUCET?

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Nishtha Mishra

Contributor-Level 7

Students have to be maximum 22 years old as on July 1, of admission year to be eligible for SUCET. If they do not meet age criteira, then they will not be allowed to participate in the admissions. Students should check official criteria before starting their registration.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Ten Exemplar Solution

Find the equation of one of the sides of an isosceles right-angled triangle whose hypotenuse is given by 3x + 4y = 4 and the opposite vertex of the hypotenuse is (2, 2).

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Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t e q u a t i o n o f t h e h y p o t e n u s e i s 3 x + 4 y = 4 a n d o p p o s i t e v e r t e x i s (2, 2) \\ S l o p e B C = \frac{- 3}{4} \\ L e t s l o p e o f A C b e m \\ ∴ t a n 4 5^{0} = | \frac{m + \frac{3}{4}}{1 + (\frac{- 3}{4}) m} | \\ \Rightarrow 1 = | \frac{4 m + 3}{4 - 3 m} | \Rightarrow \frac{4 m + 3}{4 - 3 m} = \pm 1 \\ T a k i n g (+) s i g n, \frac{4 m + 3}{4 - 3 m} = 1 \\ \Rightarrow 4 m + 3 = 4 - 3 m \\ \Rightarrow 4 m + 3 m = 4 - 3 \Rightarrow 7 m = 1 \Rightarrow m = \frac{1}{7} \\ T a k i n g (-) s i g n, \frac{4 m + 3}{4 - 3 m} = - 1 \\ \Rightarrow 4 m + 3 = - 4 + 3 m \\ \Rightarrow 4 m - 3 m = - 4 - 3 \Rightarrow m = - 7 \\ ∴ E q u a t i o n o f A C w i t h s l o p e (\frac{1}{7}) i s \\ y - 2 = \frac{1}{7} (x - 2) \\ \Rightarrow 7 y - 14 = x - 2 \Rightarrow x - 7 y + 12 = 0 \\ E q u a t i o n o f A C w i t h s l o p e (- 7) i s \\ y - 2 = - 7 (x - 2) \\ \Rightarrow y - 2 = - 7 x + 14 \\ \Rightarrow 7 x + y - 16 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n s a r e x - 7 y + 12 = 0 a n d 7 x + y - 16 = 0 . \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Nine Maths

Sol:

$\begin{array}{l} G i v e n t h a t \underset{h \to 0}{l i m} \frac{\sqrt[]{x + h} - \sqrt[]{x}}{h} \\ = \underset{h \to 0}{l i m} \frac{\sqrt[]{x + h} - \sqrt[]{x}}{h} \times \frac{\sqrt[]{x + h} + \sqrt[]{x}}{\sqrt[]{x + h} + \sqrt[]{x}} [R a t i o n a l i z i n g t h e] \\ = \underset{h \to 0}{l i m} \frac{x + h - x}{h [\sqrt[]{x + h} + \sqrt[]{x}]} = \underset{h \to 0}{l i m} \frac{h}{h [\sqrt[]{x + h} + \sqrt[]{x}]} \\ = \underset{h \to 0}{l i m} \frac{1}{\sqrt[]{x + h} + \sqrt[]{x}} \\ T a k i n g, w e h a v e \\ \frac{1}{\sqrt[]{x} + \sqrt[]{x}} = \frac{1}{2 \sqrt[]{x}} . \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Ten Exemplar Solution

Find the equation of a straight line on which the length of perpendicular from the origin is four units and the line makes an angle of 120° with the positive direction of the x-axis.

(Hint: Use normal form, where ω = 30°.)

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Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : \\ O M = 4 u n i t s \\ ∠ B A X = 12 0^{0} \\ ∴ ∠ B A O = 18 0^{0} - 12 0^{0} o r ∠ M O A + ∠ M A O = 9 0^{0} [? O M ⊥ A B] \\ θ + 6 0^{0} = 9 0^{0} \Rightarrow θ = 9 0^{0} - 6 0^{0} \\ ∴ θ = 3 0^{0} \\ S o, e q u a t i o n o f A B i n i t s n o r m a l f o r m \\ x c o s θ + y s i n θ = p \\ \Rightarrow x c o s 3 0^{0} + y s i n 3 0^{0} = 4 \\ \Rightarrow x \times \frac{\sqrt[]{3}}{2} + y \times \frac{1}{2} = 4 \Rightarrow \sqrt[]{3} x + y = 8 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s \sqrt[]{3} x + y = 8 . \end{array}$

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