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Exemplar Solution Class 11th

One vertex of the equilateral triangle with centroid at the origin and one side as x + y – 2 = 0 is

(a) (–1, –1)

(b) (2, 2)

(c) (–2, –2)

(d) (2, –2)

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Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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Exemplar Solution Class 11th

The ratio in which the line 3x + 4y + 2 = 0 divides the distance between the lines 3x + 4y + 5 = 0 and 3x + 4y – 5 = 0 is

(a) 1 : 2

(b) 3 : 7

(c) 2 : 3

(d) 2 : 5

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Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n e q u a t i o n s a r e \\ 3 x + 4 y + 5 = 0 \dots (i) \\ 3 x + 4 y - 5 = 0 \dots (i i) \\ a n d 3 x + 4 y + 2 = 0 \dots (i i i) \\ C l e a r l y, e q n . (i), (i i) a n d (i i i) a r e p a r a l l e l t o e a c h o t h e r a s t h e c o e f f i c i e n t s o f x a n d y a r e s a m e . \\ b e t w e e n p a r a l l e l l i n e s (i) a n d (i i i) w e g e t \\ = | \frac{5 - 2}{\sqrt[]{{(3)}^{2} + {(4)}^{2}}} | = \frac{3}{5} [? b e t w e e n t h e t w o p a r a l l e l l i n e s = \frac{| c_{1} - c_{2} |}{\sqrt[]{a^{2} + b^{2}}}] \\ b e t w e e n p a r a l l e l l i n e s (i i) a n d (i i i) w e g e t \\ = | \frac{- 5 - 2}{\sqrt[]{{(3)}^{2} + {(4)}^{2}}} | = \frac{7}{5} \\ ∴ R a t i o b e t w e e n t h e \frac{3}{5} : \frac{7}{5} = 3 : 7 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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Exemplar Solution Class 11th

A line passes through (2, 2) and is perpendicular to the line 3x + y = 3. Its y-intercept is

(a) $\frac{1}{3}$ 

(b) $\frac{2}{3}$  (c) 1

(d) $\frac{4}{3}$ <!-- [if gte msEquation 12]>43

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Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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Exemplar Solution Class 11th

A point equidistant from the lines 4x + 3y + 10 = 0, 5x – 12y + 26 = 0, and 7x + 24y – 50 = 0 is

(a) (1, –1)

(b) (1, 1)

(c) (0, 0)

(d) (0, 1)

3 Views|Posted a year ago

Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n e q u a t i o n s a r e \\ 4 x + 3 y + 10 = 0 \dots (i) \\ 5 x - 12 y + 26 = 0 \dots (i i) \\ a n d 7 x + 24 y - 50 = 0 \dots (i i i) \\ L e t (x_{1}, y_{1}) b e a n y f r o m e q n . (i), e q n . (i i) a n d e q n . (i i i) \\ o f (x_{1}, y_{1}) f r o m e q n . (i) \\ = | \frac{4 x_{1} + 3 y_{1} + 10}{\sqrt[]{16 + 9}} | = | \frac{4 x_{1} + 3 y_{1} + 10}{5} | \\ o f (x_{1}, y_{1}) f r o m e q n . (i i) \\ = | \frac{5 x_{1} - 12 y_{1} + 26}{\sqrt[]{25 + 144}} | = | \frac{5 x_{1} - 12 y_{1} + 26}{13} | \\ o f (x_{1}, y_{1}) f r o m e q n . (i i) \\ = | \frac{7 x_{1} + 24 y_{1} - 50}{\sqrt[]{49 + 576}} | = | \frac{7 x_{1} + 24 y_{1} - 50}{25} | \\ I f t h e (x_{1}, y_{1}) i s f r o m t h e g i v e n l i n e s, t h e n \\ | \frac{4 x_{1} + 3 y_{1} + 10}{5} | = | \frac{5 x_{1} - 12 y_{1} + 26}{13} | = | \frac{7 x_{1} + 24 y_{1} - 50}{25} | \\ W e s e e t h a t p u t t i n g x_{1} = 0 a n d y_{1} = 0, t h e a b o v e r e l a t i o n i s s a t i s f i e d i . e ., \\ \frac{10}{5} = \frac{26}{13} = \frac{50}{25} = 2 \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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Exemplar Solution Class 11th

The point (4, 1) undergoes the following two successive transformations:

i. Reflection about the line y = x

ii. Translation through a distance 2 units along the positive x-axis

Then the final coordinates of the point are

(a) (4, 3)

(b) (3, 4)

(c) (1, 4)

(d) $(\frac{7}{2}, \frac{7}{2})$ <!-- [if gte msEquation 12]>72,72

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Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t t h e r e f l e c t i o n o f A (4, 1) i n y = x b e B (a, b) m i d o f A B = (\frac{4 + a}{2}, \frac{1 + b}{2}) \\ w h i c h l i e s o n y = x \\ \Rightarrow \frac{4 + a}{2} = \frac{1 + b}{2} \\ \Rightarrow 4 + a = 1 + b \\ \Rightarrow a - b = - 3 \dots (i) \\ T h e s l o p e o f t h e l i n e y = x i s 1 a n d s l o p e o f A B = \frac{b - 1}{a - 4} \\ ∴ 1 (\frac{b - 1}{a - 4}) = - 1 \\ \Rightarrow b - 1 = - a + 4 \\ \Rightarrow a + b = 5 \dots (i i) \\ S o l v i n g e q n . (i) a n d e q n . (i i) w e g e t \\ a = 1 a n d b = 4 \\ ∴ T h e a f t e r t r a n s l a t i o n i s (1 + 2, 4) o r (3, 4) \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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Exemplar Solution Class 11th

For specifying a straight line, how many geometrical parameters should be known?

(a) 1

(b) 2

(c) 4

(d) 3

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Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} D i f f e r e n t f o r m o f e q u a t i o n o f s t r a i g h t l i n e s a r e \\ S l o p e f o r m, y = m x + c, P a r a m e t e r = 2 \\ I n t e r c e p t f o r m, \frac{x}{a} + \frac{y}{b} = 1, P a r a m e t e r = 2 \\ O n e f o r m, y - y_{1} = m (x - x_{1}), P a r a m e t e r = 2 \\ N o r m a l f o r m, x c o s w + y s i n w = P, P a r a m e t e r = 2 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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Exemplar Solution Class 11th

Equations of diagonals of the square formed by the lines x = 0, y = 0, x = 1, and y = 1 are

(a) y = x, y + x = 1

(b) y = x, x + y = 2

(c) 2y = x, y + x = $\frac{1}{3}$ (d) y = 2x, y + 2x = 1

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Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n e q u a t i o n x = 0, y = 0, x = 1 a n d y = 1 f o r m a s q u a r e o f s i d e 1 u n i t \\ F r o m f i g u r e, w e g e t t h a t O A B C i s s q u a r e h a v i n g c o r n e r s O (0, 0), A (1, 0), B (1, 1) a n d C (0, 1) \\ E q u a t i o n o f d i a g o n a l A C \\ \Rightarrow y - 0 = \frac{1 - 0}{0 - 1} (x - 1) \\ \Rightarrow y = - (x - 1) \\ \Rightarrow y = - x + 1 \\ \Rightarrow y + x = 1 \\ E q u a t i o n o f d i a g o n a l O B i s \\ \Rightarrow y - 0 = \frac{1 - 0}{1 - 0} (x - 0) \Rightarrow y = x \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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Exemplar Solution Class 11th

Equation of the line passing through (1, 2) and parallel to the line y = 3x – 1 is

(a) y + 2 = x + 1

(b) y + 2 = 3 (x + 1)

(c) y – 2 = 3 (x – 1)

(d) y – 2 = x – 1

5 Views|Posted a year ago

Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n e q u a t i o n i s y = 3 x - 1 \\ S l o p e = 3 \\ S l o p e o f t h e l i n e t h r o u g h t h e g i v e n (1, 2) a n d p a r a l l e l t o t h e g i v e n l i n e = 3 \\ S o, t h e e q u a t i o n o f t h e r e q u i r e d l i n e i s \\ \Rightarrow y - 2 = 3 (x - 1) \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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Exemplar Solution Class 11th

If the coordinates of the middle point of the portion of a line intercepted between the coordinate axes is (3, 2), then the equation of the line will be

(a) 2x + 3y = 12

(b) 3x + 2y = 12

(c) 4x – 3y = 6

(d) 5x – 2y = 10

3 Views|Posted a year ago

Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t t h e g i v e n l i n e m e e t s t h e a x e s a t A (a, 0) a n d B (0, b) . \\ G i v e n t h a t C (3, 2) i s t h e m i d o f A B \\ ∴ 3 = \frac{a + 0}{2} \Rightarrow a = 6 \\ a n d 2 = \frac{0 + b}{2} \Rightarrow b = 4 \\ I n t e r c e p t f o r m o f t h e l i n e A B \frac{x}{a} + \frac{y}{b} = 1 \\ \Rightarrow \frac{x}{6} + \frac{y}{4} = 1 \Rightarrow 2 x + 3 y = 12 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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Exemplar Solution Class 11th

The coordinates of the foot of the perpendicular from the point (2, 3) on the line $y = 3 x + 4$ is given by

(a) $(\frac{37}{10}, - \frac{1}{10})$ (b) $(- \frac{1}{10}, \frac{37}{10})$ (c) $(- \frac{10}{37}, - \frac{10}{37})$ (d) $(\frac{2}{3}, - \frac{1}{3})$  <!-- [if gte msEquation 12]>23,-13

6 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n e q u a t i o n i s y = 3 x + 4 \\ \Rightarrow 3 x - y + 4 = 0 \dots (i) \\ S l o p e = 3 \\ E q u a t i o n o f a n y l i n e t h r o u g h t h e (2, 3) i s \\ y - 3 = m (x - 2) \dots (i i) \\ I f e q n . (i) i s p e r p e n d i c u l a r t o e q n . (i i) t h e n \\ m \times 3 = - 1 [? m_{1} \times m_{2} = - 1] \\ \Rightarrow m = - \frac{1}{3} \\ P u t t i n g t h e v a l u e o f m i n e q n . (i i) w e g e t \\ \Rightarrow y - 3 = - \frac{1}{3} (x - 2) \\ \Rightarrow 3 y - 9 = - x + 2 \\ \Rightarrow x + 3 y = 11 \dots (i i i) \\ S o l v i n g e q n . (i) a n d e q n . (i i i) w e g e t \\ 3 x - y = - 4 \Rightarrow y = 3 x + 4 \dots (i v) \\ P u t t i n g t h e v a l u e o f y i n e q n . (i i i) w e g e t \\ x + 3 (3 x + 4) = 11 \\ \Rightarrow x + 9 x + 12 = 11 \\ \Rightarrow 10 x = - 1 \Rightarrow x = \frac{- 1}{10} \\ F r o m e q n . (i v) w e g e t, y = 3 (\frac{- 1}{10}) + 4 \\ \Rightarrow y = \frac{- 3}{10} + 4 \Rightarrow y = \frac{37}{10} \\ S o, t h e r e q u i r e d c o o r d i n a t e s a r e (\frac{- 1}{10}, \frac{37}{10}) . \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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One vertex of the equilateral triangle with centroid at the origin and one side as x + y – 2 = 0 is

(a) (–1, –1)

(b) (2, 2)

(c) (–2, –2)

(d) (2, –2)

The ratio in which the line 3x + 4y + 2 = 0 divides the distance between the lines 3x + 4y + 5 = 0 and 3x + 4y – 5 = 0 is

(a) 1 : 2

(b) 3 : 7

(c) 2 : 3

(d) 2 : 5

A point equidistant from the lines 4x + 3y + 10 = 0, 5x – 12y + 26 = 0, and 7x + 24y – 50 = 0 is

(a) (1, –1)

(b) (1, 1)

(c) (0, 0)

(d) (0, 1)

For specifying a straight line, how many geometrical parameters should be known?

(a) 1

(b) 2

(c) 4

(d) 3

Equation of the line passing through (1, 2) and parallel to the line y = 3x – 1 is

(a) y + 2 = x + 1

(b) y + 2 = 3 (x + 1)

(c) y – 2 = 3 (x – 1)

(d) y – 2 = x – 1