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New answer posted

a year ago

Class 11th Exemplar Solution

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

(a) 1

(b) 2

(c) 4

(d) 3

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

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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New answer posted

a year ago

Class 11th Exemplar Solution

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

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

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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New answer posted

a year ago

Class 11th Exemplar Solution

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

0 Follower 5 Views

Vishal Baghel

Contributor-Level 10

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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New answer posted

a year ago

Class 11th Exemplar Solution

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

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

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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New answer posted

a year ago

Class 11th Exemplar Solution

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) $(\frac{2}{3}, - \frac{1}{3})$

0 Follower 6 Views

Vishal Baghel

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

Class 11th Exemplar Solution

The distance between the lines $y = m x + c_{1}$ and $y = m x + c_{2}$ is

(a) $\frac{c_{1} - c_{2}}{\sqrt[]{1 + m^{2}}}$

(b) $\frac{∣ c_{1} - c_{2} ∣}{\sqrt[]{1 + m^{2}}}$

(c) $\frac{2 (c_{1} - c_{2})}{\sqrt[]{1 + m^{2}}}$

(d) 0

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

Class 11th Exemplar Solution

The equations of the lines passing through the point (1, 0) and at a distance $\frac{\sqrt[]{3}}{2}$ from the origin, are

(a) $3 x + y – 3 = 0, 3 x – y – 3 = 0$

(b) $3 x + y + 3 = 0, 3 x – y + 3 = 0$

(d) None of these

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

Class 11th Exemplar Solution

The equations of the lines which pass through the point (3, –2) and are inclined at 60° to the line 3x + y = 1 is

(a) $y + 2 = 0, 3 x – y – 2 - 3 \sqrt[]{3} = 0$

(b) $x – 2 = 0, 3 x – y + 2 + 3 \sqrt[]{3} = 0$

(c) $3 x – y – 2 - 3 \sqrt[]{3} = 0$

(d) None of these

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

Class 11th Exemplar Solution

The distance of the point of intersection of the lines 2x – 3y + 5 = 0 and 3x + 4y = 0 from the line 5x – 2y = 0 is

(a) $\frac{130}{17}$

(b) $\frac{13 \sqrt[]{7}}{29}$

(c) $\frac{130}{7}$

(d) None of these

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n e q u a t i o n s a r e : \\ 2 x - 3 y + 5 = 0 \dots (i) \\ 3 x + 4 y = 0 \dots (i i) \\ F r o m e q n . (i i) w e g e t, \\ 4 y = - 3 x \Rightarrow y = \frac{- 3}{4} x \dots (i i i) \\ P u t t i n g t h e v a l u e o f y i n e q n . (i) w e h a v e \\ 2 x - 3 (\frac{- 3}{4} x) + 5 = 0 \\ \Rightarrow 8 x + 9 x + 20 = 0 \\ \Rightarrow 17 x + 20 = 0 \Rightarrow x = \frac{- 20}{17} \\ P u t t i n g t h e v a l u e o f x i n e q n . (i i i) w e g e t \\ y = \frac{- 3}{4} (\frac{- 20}{17}) = \frac{15}{17} \\ ∴ o f i s (\frac{- 20}{17}, \frac{15}{17}) \\ N o w p e r p e n d i c u l a r f r o m t h e (\frac{- 20}{17}, \frac{15}{17}) t o t h e g i v e n l i n e 5 x - 2 y = 0 i s \\ | \frac{5 (\frac{- 20}{17}) - 2 (\frac{15}{17})}{\sqrt[]{25 + 4}} | = | \frac{\frac{- 100}{17} - \frac{30}{17}}{\sqrt[]{29}} | = \frac{130}{17 \sqrt[]{29}} \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

If the line $\frac{x}{a} + \frac{y}{b} = 1$ passes through the points (2, –3) and (4, –5), then (a, b) is

(a) (1, 1)

(b) (–1, 1)

(c) (1, –1)

(d) (–1, –1)

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} E q u a t i o n o f l i n e t h r o u g h t h e (2, - 3) a n d (4, - 5) i s \\ y + 3 = \frac{- 5 + 3}{4 - 2} (x - 2) \\ \Rightarrow y + 3 = \frac{- 2}{2} (x - 2) \\ \Rightarrow y + 3 = - (x - 2) \\ \Rightarrow y + 3 = - x + 2 \\ \Rightarrow x + y = - 1 \\ \Rightarrow \frac{x}{- 1} + \frac{y}{- 1} = 1 (I n t e r c e p t f o r m) \\ ∴ a = - 1, b = - 1 \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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