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9 months ago

GITAM School of Business, Bengaluru Colleges in Bangalore

How many students got placed during the GITAM School of Business, Bengaluru placements?

0 Follower 2 Views

Meitankeisangbam Vasundhara devi

Contributor-Level 10

In total 138 students got placed during the GITAM School of Business, Bengaluru placements 2024. Check out the table for more students placement details:

Particulars	Placement Statistics (2022)	Placement Statistics (2023)	Placement Statistics (2024)
Students Placed	106	188	138

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9 months ago

GITAM School of Business, Bengaluru Colleges in Bangalore

What is the highest package offered at GITAM School of Business, Bengaluru placements?

0 Follower 1 View

Meitankeisangbam Vasundhara devi

Contributor-Level 10

The GITAM School of Business, Bengaluru placement's the highest package stood at INR 16.75 LPA. Check out the table below to know more about the highest packages offered at during placement seasons 2022-24:

Particulars	Placement Statistics (2022)	Placement Statistics (2023)	Placement Statistics (2024)
the highest Package	INR 8.91 LPA	INR 10 LPA	INR 16.75 LPA

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Ten Exemplar Solution

The equation of the line through the intersection of the lines $2 x – 3 y = 0$ and $4 x – 5 y = 2$ and

Column C1Column C2

through the point (2, 1) is (i) $2 x – y = 4$

perpendicular to the line $x + 2 y + 1 = 0$ is (ii) $x + y – 5 = 0$

parallel to the line $3 x – 4 y + 5 = 0$ is (iii) $x – y – 1 = 0$

equally inclined to the axes is (iv) $3 x – 4 y – 1 = 0$

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

Sol:

$\begin{array}{l} (a) G i v e n ? ? e q u a t i o n s ? ? a r e ? ? 2 x ? 3 y = 0 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? \dots (i) \\ ? ? ? ? ? ? ? a n d ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 4 x ? 5 y = 2 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? \dots (i i) \\ ? ? ? ? ? ? E q u a t i o n s ? ? o f ? ? l i n e ? ? passing ? ? t h r o u g h ? ? e q n . (i) ? ? a n d ? ? (i i) ? ? w e ? ? g e t \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? (2 x ? 3 y) + k (4 x ? 5 y ? 2) = 0 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? \dots (i i i) \\ ? ? ? ? ? ? I f ? ? e q n . (i i i) ? ? p a s s e s ? ? t h r o u g h ? ? (2, 1), ? ? w e ? ? g e t \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? (2 * 2 ? 3 * 1) + k (4 * 2 ? 5 * 1 ? 2) = 0 \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? (4 ? 3) + k (8 ? 5 ? 2) = 0 \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 1 + k (8 ? 7) = 0 \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? k = ? 1 \\ ? ? ? ? ? S o, ? ? t h e ? ? r e q u i r e d ? ? e q u a t i o n ? ? i s \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? (2 x ? 3 y) ? 1 (4 x ? 5 y ? 2) = 0 \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 2 x ? 3 y ? 4 x + 5 y + 2 = 0 \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 2 x + 2 y + 2 = 0 \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? x ? y ? 1 = 0 \\ ? ? ? ? ? ? H e n c e, ? ? (a) ? (i i i) \\ (b) E q u a t i o n ? ? o f ? ? a n y ? ? l i n e ? ? passing ? ? t h r o u g h ? ? t h e ? ? point ? ? o f ? ? intersection ? ? o f ? ? t h e ? ? l i n e \\ ? ? ? ? ? ? ? 2 x ? 3 y = 0 ? ? a n d ? ? 4 x ? 5 y = 2 ? ? i s \\ ? ? ? ? ? ? ? ? ? ? (2 x ? 3 y) + k (4 x ? 5 y ? 2) = 0 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? \dots (i) \\ ? ? ? (2 + 4 k) x + (? 3 ? 5 k) y ? 2 k = 0 \\ ? ? ? ? ? ? ? ? S l o p e = \frac{? (2 + 4 k)}{? 3 ? 5 k} = \frac{2 + 4 k}{3 + 5 k} \\ ? ? ? ? ? ? ? ? S l o p e ? ? o f ? ? t h e ? ? g i v e n ? ? l i n e ? ? x + 2 y + 1 = 0 ? ? i s ? ? ? \frac{1}{2} . \\ ? ? ? ? ? ? ? ? I f ? ? t h e y ? ? a r e ? ? p e r p e n d i c u l a r ? ? t o ? ? e a c h ? ? o t h e r ? ? t h e n \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? \frac{1}{2} (\frac{2 + 4 k}{3 + 5 k}) = ? 1 \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? \frac{1 + 2 k}{3 + 5 k} = 1 \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 1 + 2 k = 3 + 5 k \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 3 k = ? 2 ? ? ? ? ? ? ? ? ? k = \frac{? 2}{3} \\ ? ? ? ? ? ? P u t t i n g ? ? t h e ? ? v a l u e ? ? o f ? ? k ? ? i n ? ? e q n . (i) ? ? w e ? ? g e t \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? (2 x ? 3 y) ? \frac{2}{3} (4 x ? 5 y ? 2) = 0 \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 6 x ? 9 y ? 8 x + 10 y + 4 = 0 \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 2 x + y + 4 = 0 \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 2 x ? y = 4 \\ ? ? ? ? ? ? H e n c e, ? ? (b) ? (i) \end{array}$

$\begin{array}{l} (c) G i v e n ? ? e q u a t i o n s ? ? a r e ? ? 2 x ? 3 y = 0 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? \dots (i) \\ ? ? ? ? ? ? ? a n d ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 4 x ? 5 y = 2 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? \dots (i i) \\ ? ? ? ? ? ? E q u a t i o n s ? ? o f ? ? l i n e ? ? passing ? ? t h r o u g h ? ? e q n . (i) ? ? a n d ? ? (i i) ? ? w e ? ? g e t \\ ? ? ? ? ? ? ? ? ? ? ? ? ? (2 x ? 3 y) + k (4 x ? 5 y ? 2) = 0 \\ ? ? ? ? ? (2 + 4 k) x + (? 3 ? 5 k) y ? 2 k = 0 \\ ? ? ? ? ? ? ? S l o p e = \frac{? (2 + 4 k)}{? 3 ? 5 k} = \frac{2 + 4 k}{3 + 5 k} \\ ? ? ? ? ? ? ? ? S l o p e ? ? o f ? ? t h e ? ? g i v e n ? ? l i n e ? ? 3 x ? 4 y + 5 = 0 ? ? i s ? ? \frac{3}{4} . \\ ? ? ? ? ? ? ? ? I f ? ? t h e ? ? t w o ? ? e q u a t i o n s ? ? a r e ? ? p a r a l l e l, ? ? t h e n \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? \frac{2 + 4 k}{3 + 5 k} = \frac{3}{4} \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 8 + 16 k = 9 + 15 k \\ ? ? ? ? ? ? ? ? ? ? ? ? ? 16 k ? 15 k = 9 ? 8 \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? k = 1 \\ ? ? ? ? ? S o, ? ? t h e ? ? r e q u i r e d ? ? e q u a t i o n ? ? o f ? ? l i n e ? ? i s \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? (2 x ? 3 y \end{array}$

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New Question 9 months ago

GITAM School of Business, Bengaluru Colleges in Bangalore

Does GITAM School of Business, Bengaluru provide 100% placements?

0 Follower 2 Views

Meitankeisangbam Vasundhara devi Contributor-Level 10

The institute has not released the placement rate as of yet. For more information on placement packages and recruiters for the GITAM School of Business, Bengaluru placements during 2022-2024, refer to the table below:

Particulars	Placement Statistics (2022)	Placement Statistics (2023)	Placement Statistics (2024)
the highest Package	INR 8.91 LPA	INR 10 LPA	INR 16.75 LPA
Average Package	INR 5.17 LPA	INR 6.21 LPA	INR 6.5 LPA
Total Recruiters	62	49	57
Students Placed	106	188	138
Total Offers	148	209	170

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New Question 9 months ago

Nightingale Institute of Management Studies Colleges in Kolkata

How does the MBA fee at NIMS Institute of Management and Commerce compare with other universities?

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Himanshu Singh Contributor-Level 10

MBA fees at NIMS Institute of Management and Commerce ranges from INR 2.8 lakh to INR 5 lakh. In comparison, MNIT Jaipur charges around INR 2 lakh, while Manipal University Jaipur has a significantly higher fee ranging from INR 10.2 lakh to INR 11.46 lakh. This comparison helps students evaluate affordability alongside course offerings and reputation.

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New Question 9 months ago

GITAM School of Business, Bengaluru GITAM School of Business, Bengaluru Placement

What is the GITAM School of Business placement truth? Does it really offer good placements?

0 Follower 3 Views

Meitankeisangbam Vasundhara devi Contributor-Level 10

The institute is known for its decent placements. For more information on the GITAM School of Business, Bengaluru placements during 2022-2024, refer to the table below:

Particulars	Placement Statistics (2022)	Placement Statistics (2023)	Placement Statistics (2024)
the highest Package	INR 8.91 LPA	INR 10 LPA	INR 16.75 LPA
Average Package	INR 5.17 LPA	INR 6.21 LPA	INR 6.5 LPA
Total Recruiters	62	49	57
Students Placed	106	188	138
Total Offers	148	209	170

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New Question 9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Ten Exemplar Solution

The equation of the line through the intersection of the lines $2 x – 3 y = 0$ and $4 x – 5 y = 2$ and

Column C1Column C2

through the point (2, 1) is (i) $2 x – y = 4$

perpendicular to the line $x + 2 y + 1 = 0$ is (ii) $x + y – 5 = 0$

parallel to the line $3 x – 4 y + 5 = 0$ is (iii) $x – y – 1 = 0$

equally inclined to the axes is (iv) $3 x – 4 y – 1 = 0$

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

Sol:

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New Question 9 months ago

GITAM School of Business, Bengaluru Colleges in Bangalore

Is placement good at GITAM School of Business, Bengaluru?

0 Follower 2 Views

Meitankeisangbam Vasundhara devi Contributor-Level 10

The GITAM School of Business is considered good when it comes to placements. For more information on placement packages and recruiters for the GITAM School of Business, Bengaluru placements during 2022-2024, refer to the table below:

Particulars	Placement Statistics (2022)	Placement Statistics (2023)	Placement Statistics (2024)
the highest Package	INR 8.91 LPA	INR 10 LPA	INR 16.75 LPA
Average Package	INR 5.17 LPA	INR 6.21 LPA	INR 6.5 LPA
Total Recruiters	62	49	57
Students Placed	106	188	138
Total Offers	148	209	170

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New Question 9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Ten Exemplar Solution

The value of $λ$ , if the lines $(2 x + 3 y + 4) + λ (6 x – y + 12) = 0$ are

Column C1 Column C2

i. Parallel to y-axis is

...more

The value of $λ$ , if the lines $(2 x + 3 y + 4) + λ (6 x – y + 12) = 0$ are

Column C1 Column C2

i. Parallel to y-axis is (i) $λ = - \frac{3}{4}$

ii. Perpendicular to $7 x + y – 4 = 0$ is (ii) $λ = - \frac{1}{3}$

iii. passes through (1, 2) is (iii) $λ = - \frac{17}{41}$

iv. parallel to x-axis is (iv) $λ = 3$

less

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Vishal Baghel

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} (a) G i v e n ? ? e q u a t i o n i s \\ (2 x + 3 y + 4) + λ (6 x - y + 12) = 0 \\ \Rightarrow (2 + 6 λ) x + (3 - λ) y + 4 + 12 λ = 0 \dots (i) \\ I f e q n . (i) i s p a r a l l e l t o y a x i s, t h e n \\ 3 - λ = 0 \Rightarrow λ = 3 \\ H e n c e, (a) \leftrightarrow (i v) \\ (b) G i v e n ? ? l i n e s a r e \\ (2 x + 3 y + 4) + λ (6 x - y + 12) = 0 \dots (i) \\ \Rightarrow (2 + 6 λ) x + (3 - λ) y + 4 + 12 λ = 0 \\ S l o p e = - (\frac{2 + 6 λ}{3 - λ}) \\ e q u a t i o n i s 7 x + y - 4 = 0 \dots (i i) \\ S l o p e = - 7 \\ I f e q n . (i) a n d e q n . (i i) a r e p e r p e n d i c u l a r t o e a c h o t h e r \\ ∴ (- 7) [- (\frac{2 + 6 λ}{3 - λ})] = - 1 \\ \Rightarrow \frac{14 + 42 λ}{3 - λ} = - 1 \\ \Rightarrow 14 + 42 λ = - 3 + λ \\ \Rightarrow 42 λ - λ = - 3 - 14 \\ \Rightarrow 41 λ = - 17 \\ \Rightarrow λ = - \frac{17}{41} \\ H e n c e, (b) \leftrightarrow (i i i) \\ (c) G i v e n ? ? e q u a t i o n i s (2 x + 3 y + 4) + λ (6 x - y + 12) = 0 \dots (i) \\ I f e q n . (i) t h r o u g h t h e g i v e n (1, 2) t h e n \\ (2 \times 1 + 3 \times 2 + 4) + λ (6 \times 1 - 2 + 12) = 0 \\ \Rightarrow (2 + 6 + 4) + λ (6 - 2 + 12) = 0 \\ \Rightarrow 12 + 16 λ = 0 \\ \Rightarrow λ = \frac{- 12}{16} = \frac{- 3}{4} \\ H e n c e, (c) \leftrightarrow (i) \\ (d) T h e g i v e n ? ? e q u a t i o n i s (2 x + 3 y + 4) + λ (6 x - y + 12) = 0 \\ \Rightarrow (2 + 6 λ) x + (3 - λ) y + 4 + 12 λ = 0 \dots (i) \\ I f e q n . (i) i s p a r a l l e l t o x a x i s, t h e n \\ 2 + 6 λ = 0 \Rightarrow λ = \frac{- 1}{3} \\ H e n c e, (d) \leftrightarrow (i i) \end{array}$

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New Question

9 months ago

GITAM School of Business, Bengaluru Colleges in Bangalore

How are the GITAM School of Business, Bengaluru placements?

0 Follower 2 Views

Meitankeisangbam Vasundhara devi

Contributor-Level 10

The institute has had a strong placement record since the very beginning. Annually, leading companies take part in placing graduating students. For more information on placement packages and recruiters for the GITAM School of Business, Bengaluru placements for 2022-2024, refer to the table below:

Particulars	Placement Statistics (2022)	Placement Statistics (2023)	Placement Statistics (2024)
the highest Package	INR 8.91 LPA	INR 10 LPA	INR 16.75 LPA
Average Package	INR 5.17 LPA	INR 6.21 LPA	INR 6.5 LPA
Total Recruiters	62	49	57
Students Placed	106	188	138
Total Offers	148	209	170

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9 months ago

Study Abroad France

Which is better for studies: France or Germany?

0 Follower 4 Views

Akash Gaur

Contributor-Level 10

Both France & Germany are among the popular destinations for higher studies abroad. While Frennch universities are known for following a more theoretical approach to studies, German institutes lay a major emphasis on practical implementation of concepts.

The table below compares the two countries based on some key parameters, & would allow students assess the pros & cons of both countries:

Parameters	France	Germany
Popular Programmes	Business Studies, Engineering, Goods Management, Fashion Designing	Engineering, Business & Sciences
Language of Instruction	French is the first choice language, but many universities teach in English too	German is first choice, however, the number of English taught courses are growing
University Rankings	20 French universities feature in the QS World University Rankings 2026.	17 German universities feature in the QS World University Rankings 2026.
Teaching Approach	Follow a more theatrical approach Encourages critical thinking & independent research	Lays a major emphasis on practical implementation of concepts. Group projects, case studies, and internships, a key part of the academic curriculum.
Cost of Studying	Low cost of studies in the Public universities Private Universities however, can be costly Read: Cost of Studying in France	Public universities in Germany offer education at negligible tuition costs (a nominal administrative fee is charged every semester) Read: Cost of Studying in Germany
Cost of Living	Living in France is generally more expensive than Germany. Prominent cities like Paris & Lyon have higher costs of living & food.	Living Costs in Germany are generally more affordable than France. Cities like Munich are relatively more expensive, but towns offer affordable housing.
Part-Time Work Options	Students can work part-time up to 964 hours per year. Language barriers can pose a major challenge when searching for part-time work.	International students are allowed to part-time in Germany for up to 120 full days or 2,880 hours a year.
Post Study Work Permit	France offers a Temporary Residence Permit (titre de séjour) valid for up to 18 months post graduation.	Germany too offers a post-study work visa, valid for upto 18 months.
Job Market	Automotive & Aerospace engineering are among the popular industries, along with IT & Business Analytics. Laearning French however, is crucial for finding employment.	Automobile Engineering, IT & Manufacturing among the popular job markets.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Ten Exemplar Solution

Match the following:

Column C1 Column C2

(a) The coordinates of the points P and Q on the (i) $(3, 1), (– 7, 11)$

line $x + 5 y = 13$ , which are at a distance of 2 units

from the line $12 x – 5 y + 26 = 0$ are

(b) The coordinates of the point on the (ii) $(3, 1), (– 7, 11)$

line $x + y = 4$ , which is at a unit distance

from the line $4 x + 3 y – 10 = 0$ .

joining A(–2, 5) and B(3, 1) such that

$A P = P Q = Q B$ are

less

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} (a) L e t P (x_{1}, y_{1}) b e a n y o n t h e g i v e n l i n e \\ x + 5 y = 13 ∴ x_{1} + 5 y_{1} = 13 \\ o f l i n e 12 x - 5 y + 26 = 0 f r o m t h e (x_{1}, y_{1}) \\ 2 = | \frac{12 x_{1} - 5 y_{1} + 26}{\sqrt[]{{(12)}^{2} + {(- 5)}^{2}}} | \Rightarrow 2 = | \frac{12 x_{1} - (13 - x_{1}) + 26}{13} | \\ \Rightarrow 2 = | \frac{12 x_{1} - 13 + x_{1} + 26}{13} | \Rightarrow 2 = | \frac{13 x_{1} + 13}{13} | \\ \Rightarrow 2 = \pm (x_{1} + 1) \\ \Rightarrow 2 = x_{1} + 1 \Rightarrow x_{1} = 1 (T a k i n g (+) s i g n) \\ a n d 2 = - x_{1} - 1 \Rightarrow x_{1} = - 3 (T a k i n g (-) s i g n) \\ P u t t i n g t h e v a l u e o f x_{1} i n e q n . x_{1} + 5 y_{1} = 13 \\ W e g e t y_{1} = \frac{12}{5} a n d \frac{16}{5} \\ S o, t h e r e q u i r e d a r e (1, \frac{12}{5}) a n d (- 3, \frac{16}{5}) . \\ H e n c e, (a) \leftrightarrow (i i i) \\ (b) L e t P (x_{1}, y_{1}) b e a n y o n t h e g i v e n l i n e \\ x + y = 4 ∴ x_{1} + y_{1} = 4 \dots (i) \\ o f l i n e 4 x + 3 y - 10 = 0 f r o m t h e (x_{1}, y_{1}) \\ 1 = | \frac{4 x_{1} + 3 y_{1} - 10}{\sqrt[]{{(4)}^{2} + {(3)}^{2}}} | \Rightarrow 1 = | \frac{4 x_{1} + 3 (4 - x_{1}) - 10}{5} | \\ \Rightarrow 1 = | \frac{4 x_{1} + 12 - 3 x_{1} - 10}{5} | \Rightarrow 1 = | \frac{x_{1} + 2}{5} | \\ \Rightarrow 1 = \pm (\frac{x_{1} + 2}{5}) \\ \Rightarrow \frac{x_{1} + 2}{5} = 1 (T a k i n g (+) s i g n) \\ \Rightarrow x_{1} + 2 = 5 \Rightarrow x_{1} = 3 \\ a n d \frac{x_{1} + 2}{5} = - 1 (T a k i n g (-) s i g n) \\ \Rightarrow x_{1} + 2 = - 5 \Rightarrow x_{1} = - 7 \\ P u t t i n g t h e v a l u e o f x_{1} i n e q n . (i) w e g e t \\ A t x_{1} = 3, y_{1} = 1 \\ A t x_{1} = - 7, y_{1} = 11 \\ S o, t h e r e q u i r e d a r e (3, 1) a n d (- 7, 11) . \\ H e n c e, (b) \leftrightarrow (i) . \end{array}$

$\begin{array}{l} (c) G i v e n t h a t A P = P Q = Q B \\ E q u a t i o n o f l i n e j o i n i n g A (- 2, 5) a n d B (3, 1) i s \\ y - 5 = \frac{1 - 5}{3 + 2} (x + 2) \\ \Rightarrow y - 5 = \frac{- 4}{5} (x + 2) \\ \Rightarrow 5 y - 25 = - 4 x - 8 \\ \Rightarrow 4 x + 5 y - 17 = 0 \\ L e t P (x_{1}, y_{1}) a n d Q (x_{2}, y_{2}) b e a n y t w o points o n t h e l i n e A B \\ P (x_{1}, y_{1}) d i v i d e s t h e l i n e A B i n t h e r a t i o 1 : 2 \\ ∴ x_{1} = \frac{1.3 + 2 (- 2)}{1 + 2} = \frac{3 - 4}{3} = \frac{- 1}{3} \\ y_{1} = \frac{1.1 + 2.5}{1 + 2} = \frac{1 + 10}{3} = \frac{11}{3} \\ S o, t h e c o o r d i n a t e s o f P (x_{1}, y_{1}) = (\frac{- 1}{3}, \frac{11}{3}) . \\ N o w point Q (x_{2}, y_{2}) i s t h e m i d -point o f P B \\ ∴ x_{2} = \frac{3 - \frac{1}{3}}{2} = \frac{4}{3} \\ y_{2} = \frac{1 + \frac{11}{3}}{2} = \frac{7}{3} \\ H e n c e, t h e c o o r d i n a t e s o f Q (x_{2}, y_{2}) = (\frac{4}{3}, \frac{7}{3}) . \\ H e n c e, (c) \leftrightarrow (i i) . \end{array}$

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New Question

9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Ten Exemplar Solution

The line joining the points (3, –4) and (–2, 6) is perpendicular to the line joining the points (–3, 6) and (9, –18).

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} T h e g i v e n a r e (3, - 4) a n d (- 2, 6), (- 3, 6) a n d (9, - 18) . \\ S l o p e o f t h e l i n e j o i n i n g t h e (3, - 4) a n d (- 2, 6) \\ m_{1} = \frac{6 + 4}{- 2 - 3} = \frac{10}{- 5} = - 2 \\ S l o p e o f t h e l i n e j o i n i n g t h e (- 3, 6) a n d (9, - 18) \\ m_{2} = \frac{- 18 - 6}{9 + 3} = \frac{- 24}{12} = - 2 \\ S i n c e m_{1} = m_{2} = - 2 \\ S o, t h e l i n e s a r e p a r a l l e l a n d n o t p e r p e n d i c u l a r . \\ H e n c e t h e g i v e n s t a t e m e n t i s F a l s e . \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Ten Exemplar Solution

The lines $a x + 2 y + 1 = 0, b x + 3 y + 1 = 0,$ and $c x + 4 y + 1 = 0$ are concurrent if a, b, c are in G.P.

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n e q u a t i o n s a r e \\ a x + 2 y + 1 = 0 \dots (i) \\ b x + 3 y + 1 = 0 \dots (i i) \\ c x + 4 y + 1 = 0 \dots (i i i) \\ S o l v i n g e q n . (i) a n d (i i) w e g e t \\ a x + 2 y + 1 = 0 \Rightarrow y = \frac{- a x - 1}{2} \\ P u t t i n g t h e v a l u e o f y i n e q n . (i i) w e h a v e \\ b x + 3 (\frac{- a x - 1}{2}) + 1 = 0 \\ \Rightarrow 2 b x - 3 a x - 3 + 2 = 0 \\ \Rightarrow (2 b - 3 a) x = 1 \\ \Rightarrow x = \frac{1}{2 b - 3 a} \\ ∴ y = \frac{- a (\frac{1}{2 b - 3 a}) - 1}{2} \\ = \frac{- a - 2 b + 3 a}{2 (2 b - 3 a)} = \frac{2 a - 2 b}{2 (2 b - 3 a)} = \frac{a - b}{2 b - 3 a} \\ S o, t h e o f o f e q n . (i) a n d (i i) i s \\ (\frac{1}{2 b - 3 a}, \frac{a - b}{2 b - 3 a}) \\ I f e q n . (i), (i i) a n d (i i i) a r e c o n c u r r e n t, t h e n t h e a b o v e m u s t l i e o n e q n . (i i i) \\ c x + 4 y + 1 = 0 \\ \Rightarrow c [\frac{1}{2 b - 3 a}] + 4 [\frac{a - b}{2 b - 3 a}] + 1 = 0 \\ \Rightarrow \frac{c + 4 a - 4 b + 2 b - 3 a}{2 b - 3 a} = 0 \\ \Rightarrow c + a - 2 b = 0 \\ \Rightarrow 2 b = a + c \\ S o, a, b a n d c a r e i n A . P . a n d n o t i n G . P . \\ H e n c e t h e g i v e n s t a t e m e n t i s F a l s e . \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Ten Exemplar Solution

The line $\frac{x}{a} + \frac{y}{b} = 1$ moves in such a way that $\frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{c^{2}}$ , where c is a constant. The locus of the foot of the perpendicular from the origin on the given line is $x^{2} + y^{2} = c^{2}$ .

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Vishal Baghel

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} T h e g i v e n e q u a t i o n i s \frac{x}{a} + \frac{y}{b} = 1 \dots (i) \\ E q u a t i o n o f l i n e t h r o u g h t h e (0, 0) a n d p e r p e n d i c u l a r t o e q n . (i) i s \\ \frac{x}{b} - \frac{y}{a} = 0 \dots (i i) \\ S q u a r i n g a n d a d d i n g e q n . (i) a n d (i i) w e g e t \\ {(\frac{x}{a} + \frac{y}{b})}^{2} + {(\frac{x}{b} - \frac{y}{a})}^{2} = 1 + 0 \\ \Rightarrow \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{2 x y}{a b} + \frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} - \frac{2 x y}{a b} = 1 \\ \Rightarrow x^{2} (\frac{1}{a^{2}} + \frac{1}{b^{2}}) + y^{2} (\frac{1}{b^{2}} + \frac{1}{a^{2}}) = 1 \\ \Rightarrow (x^{2} + y^{2}) (\frac{1}{a^{2}} + \frac{1}{b^{2}}) = 1 \\ \Rightarrow (x^{2} + y^{2}) (\frac{1}{c^{2}}) = 1 [? \frac{1}{c^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}}] \\ \Rightarrow x^{2} + y^{2} = c^{2} \\ H e n c e t h e g i v e n s t a t e m e n t i s T r u e . \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Ten Exemplar Solution

The equation of the line joining the point (3, 5) to the point of intersection of the lines 4x + y – 1 = 0 and 7x – 3y – 35 = 0 is equidistant from the points (0, 0) and (8, 34).

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Vishal Baghel

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n e q u a t i o n s a r e \\ 4 x + y - 1 = 0 \dots (i) \\ a n d 7 x - 3 y - 35 = 0 \dots (i i) \\ F r o m e q n . (i) y = 1 - 4 x \dots (i i i) \\ P u t t i n g t h e v a l u e o f y i n e q n . (i i) w e g e t \\ 7 x - 3 (1 - 4 x) - 35 = 0 \\ \Rightarrow 7 x - 3 + 12 x - 35 = 0 \\ \Rightarrow 19 x - 38 = 0 \\ \Rightarrow x = 2 \\ F r o m e q n . (i i i) y = 1 - 4 \times 2 \Rightarrow y = - 7 \\ T h e o f i s (2, - 7) . \\ E q u a t i o n o f l i n e j o i n i n g t h e (3, 5) t o t h e (2, - 7) i s \\ y - 5 = \frac{- 7 - 5}{2 - 3} (x - 3) \\ \Rightarrow y - 5 = 12 (x - 3) \\ \Rightarrow y - 5 = 12 x - 36 \\ \Rightarrow 12 x - y - 31 = 0 \dots (i v) \\ o f e q n . (i v) f r o m t h e (0, 0) = | \frac{- 31}{\sqrt[]{{(12)}^{2} + {(- 1)}^{2}}} | = \frac{31}{\sqrt[]{145}} \\ o f e q n . (i v) f r o m t h e (8, 34) = | \frac{12 \times 8 - 34 - 31}{\sqrt[]{{(12)}^{2} + {(- 1)}^{2}}} | = | \frac{96 - 65}{\sqrt[]{145}} | = \frac{31}{\sqrt[]{145}} \\ H e n c e t h e g i v e n s t a t e m e n t i s T r u e . \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Ten Exemplar Solution

The vertex of an equilateral triangle is (2, 3) and the equation of the opposite side is x + y = 2. Then the other two sides are $y – 3 = (2 \pm \sqrt[]{3}) (x – 2)$ .

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Vishal Baghel

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t A B C b e a n e q u i l a t e r a l t r i a n g l e w i t h v e r t e x (2, 3) a n d t h e o p p o s i t e s i d e i s x + y = 2 \\ w i t h s l o p e - 1 . S u p p o s e s l o p e o f l i n e A B i s m . \\ e a c h a n g l e o f e q u i l a t e r a l t r i a n g l e i s 6 0^{0} . \\ ∴ A n g l e b e t w e e n A B a n d B C \\ t a n 6 0^{0} = | \frac{- 1 - m}{1 + (- 1) m} | \\ \Rightarrow \sqrt[]{3} = | \frac{1 + m}{1 - m} | \\ \Rightarrow \sqrt[]{3} = \pm (\frac{1 + m}{1 - m}) \\ T a k i n g (+) s i g n \\ \Rightarrow \sqrt[]{3} = \frac{1 + m}{1 - m} \Rightarrow \sqrt[]{3} - \sqrt[]{3} m = 1 + m \\ \Rightarrow \sqrt[]{3} m + m = \sqrt[]{3} - 1 \Rightarrow m (\sqrt[]{3} + 1) = \sqrt[]{3} - 1 \\ \Rightarrow m = \frac{\sqrt[]{3} - 1}{\sqrt[]{3} + 1} \Rightarrow m = \frac{\sqrt[]{3} - 1}{\sqrt[]{3} + 1} \times \frac{\sqrt[]{3} - 1}{\sqrt[]{3} - 1} \\ \Rightarrow m = \frac{3 + 1 - 2 \sqrt[]{3}}{3 - 1} = 2 - \sqrt[]{3} \\ T a k i n g (-) s i g n \\ \Rightarrow \sqrt[]{3} = - (\frac{1 + m}{1 - m}) \Rightarrow \sqrt[]{3} - \sqrt[]{3} m = - 1 - m \\ \Rightarrow - \sqrt[]{3} m + m = - 1 - \sqrt[]{3} \Rightarrow m (- \sqrt[]{3} + 1) = - 1 - \sqrt[]{3} \\ \Rightarrow m = \frac{- 1 - \sqrt[]{3}}{1 - \sqrt[]{3}} \Rightarrow m = \frac{- 1 - \sqrt[]{3}}{1 - \sqrt[]{3}} \times \frac{1 + \sqrt[]{3}}{1 + \sqrt[]{3}} \\ \Rightarrow m = \frac{- 1 - \sqrt[]{3} - \sqrt[]{3} - 3}{1 - 3} = \frac{- 4 - 2 \sqrt[]{3}}{- 2} = \frac{- 2 (2 + \sqrt[]{3})}{- 2} = 2 + \sqrt[]{3} \\ S o, t h e e q u a t i o n s o f o t h e r t w o l i n e s a r e \\ y - 3 = (2 \pm \sqrt[]{3}) (x - 2) \\ H e n c e, t h e g i v e n s t a t e m e n t i s T r u e . \end{array}$

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9 months ago

Chhattisgarh Bhilai

What is the cut off of IIT Bhilai?

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Manpreet Saxena

Contributor-Level 7

IIT Bhilai JEE Advanced cutoff 2025 has been released for round 3. The cutoff was released for different categories belonging to the All India quota. The cutoff for the General AI category candidates varied between 6524 and 15325. The lowest cutoff was for admission to BTech in CSE, and the highest cutoff was for admission to B.Tech. in Materials Science and Metallurgical Engineering.

For the OBC AI category candidates, the cutoff ranks in the third round varied between 2737 and 5744. Candidates can further check the category-wise and round-wise cutoff ranks on this page.

Branch-wise cutoff for round 3, for the General AI category

...more

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Ten Exemplar Solution

The straight line 5x + 4y = 0 passes through the point of intersection of the straight lines x + 2y – 10 = 0 and 2x + y + 5 = 0.

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Vishal Baghel

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n e q u a t i o n s a r e x + 2 y - 10 = 0 \dots (i) \\ a n d 2 x + y + 5 = 0 \dots (i i) \\ F r o m e q n . (i) x = 10 - 2 y \dots (i i i) \\ P u t t i n g t h e v a l u e o f x i n e q n . (i i) w e g e t \\ 2 (10 - 2 y) + y + 5 = 0 \\ \Rightarrow 20 - 4 y + y + 5 = 0 \\ \Rightarrow - 3 y + 25 = 0 \\ \Rightarrow y = \frac{25}{3} \\ P u t t i n g t h e v a l u e o f y i n e q n . (i i i) w e g e t \\ x = 10 - 2 (\frac{25}{3}) \\ = \frac{30 - 50}{3} = \frac{- 20}{3} \\ ∴ = (\frac{- 20}{3}, \frac{25}{3}) \\ I f t h e g i v e n l i n e 5 x + 4 y = 0 t h r o u g h t h e (\frac{- 20}{3}, \frac{25}{3}) t h e n \\ 5 (\frac{- 20}{3}) + 4 (\frac{25}{3}) = 0 \\ \Rightarrow \frac{- 100}{3} + \frac{100}{3} = 0 \\ \Rightarrow 0 = 0 s a t i s f i e d . \\ S o, t h e g i v e n l i n e t h r o u g h t h e o f o f t h e g i v e n l i n e s . \\ H e n c e, t h e g i v e n s t a t e m e n t i s T r u e . \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Ten Exemplar Solution

Equation of the line passing through the point $(a {c o s}^{3} θ, a {s i n}^{3} θ)$ and perpendicular to the line $x s e c θ + y c o s e c θ = a$ is $x c o s θ – y s i n θ = a s i n 2 θ$ .

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Vishal Baghel

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} E q u a t i o n o f a n y l i n e p e r p e n d i c u l a r t o x s e c θ + y c o s e c θ = a i s \\ x c o s e c θ - y s e c θ = k \dots (i) \\ I f e q n . (i) p a s s e s t h r o u g h (a c o s^{3} θ, a s i n^{3} θ) t h e n \\ a c o s^{3} θ . c o s e c θ - a s i n^{3} θ . s e c θ = k \\ \Rightarrow \frac{a c o s^{3} θ}{s i n θ} - \frac{a s i n^{3} θ}{c o s θ} = k \\ ∴ e q u a t i o n i s \\ x c o s e c θ - y s e c θ = \frac{a c o s^{3} θ}{s i n θ} - \frac{a s i n^{3} θ}{c o s θ} \\ \Rightarrow \frac{x}{s i n θ} - \frac{y}{c o s θ} = a [\frac{c o s^{4} θ - s i n^{4} θ}{s i n θ c o s θ}] \\ \Rightarrow \frac{x c o s θ - y s i n θ}{s i n θ c o s θ} = a [\frac{(c o s^{2} θ + s i n^{2} θ) (c o s^{2} θ - s i n^{2} θ)}{s i n θ c o s θ}] \\ \Rightarrow x c o s θ - y s i n θ = a (c o s^{2} θ - s i n^{2} θ) \\ \Rightarrow x c o s θ - y s i n θ = a c o s 2 θ \\ S o, t h e g i v e n s t a t e m e n t i s F a l s e . \end{array}$

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