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Class 11th

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New answer posted

a year ago

Class 11th Exemplar Solution

Find the equation of the line passing through the point of intersection of 2x + y = 5 and x + 3y + 8 = 0 and parallel to the line 3x + 4y = 7.

0 Follower 6 Views

Vishal Baghel

Contributor-Level 10

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

Sol:

$(a) G i v e n$

$\begin{array}{l} t h a t : 2 x + y = 5 \dots (i) \\ x + 3 y + 8 = 0 \dots (i i) \\ 3 x + 4 y = 7 \dots (i i i) \\ E q u a t i o n o f a n y l i n e t h r o u g h t h e o f o f e q n . (i) a n d e q n . (i i) i s \\ (2 x + y - 5) + λ (x + 3 y + 8) = 0 \dots (i v) [λ =] \\ \Rightarrow 2 x + y - 5 + λ x + 3 λ y + 8 λ = 0 \\ \Rightarrow (2 + λ) x + (1 + 3 λ) y - 5 + 8 λ = 0 \\ S l o p e o f l i n e m_{1} (s a y) = - \frac{(2 + λ)}{1 + 3 λ} [? m = \frac{- a}{b}] \\ S l o p e o f l i n e 3 x + 4 y = 7 i s \\ m_{2} (s a y) = - \frac{3}{4} \\ I f e q n . (i i i) i s p a r a l l e l t o e q n . (i v) t h e n \\ m_{1} = m_{2} \\ ∴ - \frac{(2 + λ)}{1 + 3 λ} = - \frac{3}{4} \\ \Rightarrow \frac{2 + λ}{1 + 3 λ} = \frac{3}{4} \Rightarrow 8 + 4 λ = 3 + 9 λ \\ \Rightarrow 9 λ - 4 λ = 5 \Rightarrow 5 λ = 5 \Rightarrow λ = 1 \\ O n p u t t i n g t h e v a l u e o f λ i n e q n . (i v) w e g e t \\ (2 x + y - 5) + 1 (x + 3 y + 8) = 0 \\ \Rightarrow 2 x + y - 5 + x + 3 y + 8 = 0 \\ \Rightarrow 3 x + 4 y + 3 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s 3 x + 4 y + 3 = 0 . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

Find the equation of lines passing through (1, 2) and making an angle 30° with the y-axis.

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t t h e l i n e m a k e s a n g l e 3 0^{0} w i t h y a x i s \\ ∴ A n g l e m a d e b y t h e l i n e w i t h x a x i s i s 6 0^{0} \\ ∴ S l o p e o f t h e l i n e \\ m = t a n 6 0^{0} \\ \Rightarrow m = \sqrt[]{3} \\ S o, t h e e q u a t i o n o f t h e l i n e t h r o u g h t h e (1, 2) a n d s l o p e \sqrt[]{3} i s \\ y - y_{1} = m (x - x_{1}) \\ \Rightarrow y - 2 = \sqrt[]{3} (x - 1) \\ \Rightarrow y - 2 = \sqrt[]{3} x - \sqrt[]{3} \\ \Rightarrow y - \sqrt[]{3} x - \sqrt[]{3} - 2 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n o f l i n e i s y - \sqrt[]{3} x - \sqrt[]{3} - 2 = 0 . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

Show that the tangent of an angle between the lines x/a + y/b = 1 and x/a − y/b = 1 is 2ab / (a² − b²).

0 Follower 6 Views

Vishal Baghel

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t : \frac{x}{a} + \frac{y}{b} = 1 \dots (i) \\ a n d \frac{x}{a} - \frac{y}{b} = 1 \dots (i i) \\ S l o p e o f t h e e q n . (i) m_{1} (s a y) = - \frac{b}{a} \\ a n d s l o p e o f t h e e q n . (i i) m_{2} (s a y) = \frac{b}{a} \\ L e t θ b e t h e a n g l e b e t q w e e n t h e t w o g i v e n l i n e s \\ ∴ t a n θ = | \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} | = | \frac{- \frac{b}{a} - \frac{b}{a}}{1 + (- \frac{b}{a}) (\frac{b}{a})} | = | \frac{\frac{- 2 b}{a}}{1 - \frac{b^{2}}{a^{2}}} | = | \frac{- 2 a b}{a^{2} - b^{2}} | \\ \Rightarrow t a n θ = \frac{2 a b}{a^{2} - b^{2}} . H e n c e p r o v e d . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

Find the points on the line x + y = 4 which lie at a unit distance from the line 4x + 3y = 10.

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

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

Sol:

$\begin{array}{l} L e t (x_{1}, y_{1}) b e a n y l y i n g i n t h e e q u a t i o n x + y = 4 \\ ∴ x_{1} + y_{1} = 4 \dots (i) \\ o f t h e (x_{1}, y_{1}) f r o m t h e e q u a t i o n 4 x + 3 y = 10 \\ \frac{4 x_{1} + 3 y_{1} - 10}{\sqrt[]{{(4)}^{2} + {(3)}^{2}}} = 1 \\ | \frac{4 x_{1} + 3 y_{1} - 10}{5} | = 1 \\ 4 x_{1} + 3 y_{1} - 10 = \pm 5 \\ T a k i n g (+) s i g n 4 x_{1} + 3 y_{1} - 10 = 5 \\ \Rightarrow 4 x_{1} + 3 y_{1} = 15 \dots (i i) \\ F r o m e q n . (i) w e g e t y_{1} = 4 - x_{1} \\ P u t t i n g t h e v a l u e o f y_{1} i n e q n . (i i) w e g e t \\ 4 x_{1} + 3 (4 - x_{1}) = 15 \\ \Rightarrow 4 x_{1} + 12 - 3 x_{1} = 15 \\ \Rightarrow x_{1} + 12 = 15 \\ \Rightarrow x_{1} = 3 a n d y_{1} = 4 - 3 = 1 \\ S o, t h e r e q u i r e d i s (3, 1) \\ N o w t a k i n g (-) s i g n, 4 x_{1} + 3 y_{1} - 10 = - 5 \\ \Rightarrow 4 x_{1} + 3 y_{1} = 5 \dots (i i i) \\ F r o m e q n . (i) w e g e t y_{1} = 4 - x_{1} \\ 4 x_{1} + 3 (4 - x_{1}) = 5 \\ \Rightarrow 4 x_{1} + 12 - 3 x_{1} = 5 \\ \Rightarrow x_{1} + 12 = 5 \\ \Rightarrow x_{1} = - 7 a n d y_{1} = 4 - (- 7) = 11 \\ S o, t h e r e q u i r e d i s (- 7, 11) \\ H e n c e, t h e r e q u i r e d o n t h e g i v e n l i n e a r e (3, 1) a n d (- 7, 11) . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

Find the equation of the lines which passes through the point (3, 4) and cuts off intercepts from the coordinate axes such that their sum is 14.

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

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

Sol:

$\begin{array}{l} E q u a t i o n o f l i n e h a v i n g a a n d b o n t h e a x i s i s \\ \frac{x}{a} + \frac{y}{b} = 1 \dots (i) \\ G i v e n t h a t a + b = 14 \Rightarrow b = 14 - a \\ \Rightarrow \frac{x}{a} + \frac{y}{14 - a} = 1 \dots (i i) \\ I f e q n . (i i) p a s s e s t h r o u g h t h e (3, 4) t h e n \\ \frac{3}{a} + \frac{4}{14 - a} = 1 \\ \Rightarrow \frac{3 (14 - a) + 4 a}{a (14 - a)} = 1 \\ \Rightarrow 42 + a = 14 a - a^{2} \\ \Rightarrow a^{2} + a - 14 a + 42 = 0 \\ \Rightarrow a^{2} - 13 a + 42 = 0 \\ \Rightarrow a^{2} - 7 a - 6 a + 42 = 0 \\ \Rightarrow a (a - 7) - 6 (a - 7) = 0 \\ \Rightarrow (a - 6) (a - 7) = 0 \Rightarrow a = 6, 7 \\ ∴ b = 14 - 6 = 8, b = 14 - 7 = 7 \\ H e n c e, t h e r e q u i r e d e q u a t i o n o f l i n e s a r e \\ \frac{x}{6} + \frac{y}{8} = 1 \Rightarrow 4 x + 3 y = 24 \\ a n d \frac{x}{7} + \frac{y}{7} = 1 \Rightarrow x + y = 7 \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

Find the equation of the circle which touches the $x$ -axis and whose center is $(1, 2)$ .

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} S i n c e t h e c i r c l e w h o s e c e n t r e i s (1, 2) t o u c h x - a x i s \\ ∴ r = 2 \\ S o, t h e e q u a t i o n o f t h e c i r c l e i s \\ {(x - h)}^{2} + {(y - k)}^{2} = r^{2} \\ \Rightarrow {(x - 1)}^{2} + {(y - 2)}^{2} = {(2)}^{2} \\ \Rightarrow x^{2} - 2 x + 1 + y^{2} - 4 y + 4 = 4 \\ \Rightarrow x^{2} + y^{2} - 2 x - 4 y + 1 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s x^{2} + y^{2} - 2 x - 4 y + 1 = 0 . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

Find the angle between the lines y = (2 – 3)(x + 5) and y = (2 + 3)(x – 7).

0 Follower 5 Views

Vishal Baghel

Contributor-Level 10

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

Sol:

$\begin{array}{l} T h e g i v e n e q u a t i o n s a r e y = (2 - \sqrt[]{3}) (x + 5) \dots (i) \\ a n d y = (2 + \sqrt[]{3}) (x - 7) \dots (i i) \\ S l o p e o f t h e e q n . (i) m_{1} (s a y) = (2 - \sqrt[]{3}) \\ a n d s l o p e o f t h e e q n . (i i) m_{2} (s a y) = (2 + \sqrt[]{3}) \\ L e t θ b e t h e a n g l e b e t q w e e n t h e t w o g i v e n l i n e s \\ ∴ t a n θ = | \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} | = | \frac{2 - \sqrt[]{3} - 2 - \sqrt[]{3}}{1 + (2 - \sqrt[]{3}) (2 + \sqrt[]{3})} | = | \frac{- 2 \sqrt[]{3}}{2} | = | - \sqrt[]{3} | \\ \Rightarrow t a n θ = \sqrt[]{3} o r - \sqrt[]{3} \\ ∴ θ = 6 0^{0} o r 12 0^{0} \\ H e n c e, t h e r e q u i r e d a n g l e i s 6 0^{0} o r 12 0^{0} . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

If a circle passes through the points $(0, 0), (a, 0), (0, b)$ , then find the coordinates of its center.

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n points a r e (0, 0), (a, 0) a n d (0, b) \\ G e n e r a l e q u a t i o n o f t h e c i r c l e i s \\ x^{2} + y^{2} + 2 g x + 2 f y + c = 0 w h e r e t h e c e n t r e i s (- g, - f) a n d r a d i u s = \sqrt[]{g^{2} + f^{2} - c} \\ I f i t p a s s e s t h r o u g h (0, 0) \\ ∴ c = 0 \\ I f i t p a s s e s t h r o u g h (a, 0) a n d (0, b) t h e n \\ a^{2} + 2 g a + c = 0 \Rightarrow a^{2} + 2 g a = 0 [? c = 0] \\ ∴ g = - \frac{a}{2} \\ a n d 0 + b^{2} + 0 + 2 f b + c = 0 \Rightarrow b^{2} + 2 f b = 0 [? c = 0] \\ \Rightarrow f = - \frac{b}{2} \\ H e n c e, t h e c o o r d i n a t e s o f c e n t r e o f t h e c i r c l e a r e (- g, - f) = (\frac{a}{2}, \frac{b}{2}) \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

Find the equation of the line passing through the point (5, 2) and perpendicular to the line joining the points (2, 3) and (3, –1).

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

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

Sol:

$S l o p e o f t h e l i n e j o i n i n g t h r o u g h t h e (2, 3) a n d (3, - 1) i s \frac{- 1 - 3}{3 - 2} = - 4 S l o p e o f t h e r e q u i r e d l i n e w h i c h i s p e r p e n d i c u l a r t o i t = \frac{- 1}{- 4} = \frac{1}{4} [$
$m_{1} m_{2} = - 1] E q u a t i o n o f t h e l i n e t h r o u g h t h e (5, 2) i s y - 2 = \frac{1}{4} (x - 5) [$
$\begin{array}{l} y - y_{1} = m (x - x_{1})] \\ \Rightarrow 4 y - 8 = x - 5 \\ \Rightarrow x - 4 y + 3 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s x - 4 y + 3 = 0 . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

Show that the point $(x, y)$ given by $x = \frac{2 a t}{1 + t^{2}}$ and $y = \frac{a (1 - t^{2})}{1 + t^{2}}$ lies on a circle for all real values of $t$ , such that $- 1 < t < 1$ , where $a$ is any given real number.

0 Follower 5 Views

Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : x = \frac{2 a t}{1 + t^{2}} a n d y = \frac{a (1 - t^{2})}{1 + t^{2}} \\ \Rightarrow x^{2} + y^{2} = {(\frac{2 a t}{1 + t^{2}})}^{2} + {(\frac{a (1 - t^{2})}{1 + t^{2}})}^{2} \\ = \frac{4 a^{2} t^{2}}{{(1 + t^{2})}^{2}} + \frac{a^{2} {(1 - t^{2})}^{2}}{{(1 + t^{2})}^{2}} = \frac{4 a^{2} t^{2} + a^{2} (1 + t^{4} - 2 t^{2})}{{(1 + t^{2})}^{2}} \\ = \frac{4 a^{2} t^{2} + a^{2} + a^{2} t^{4} - 2 a^{2} t^{2}}{{(1 + t^{2})}^{2}} = \frac{a^{2} + a^{2} t^{4} + 2 a^{2} t^{2}}{{(1 + t^{2})}^{2}} \\ = \frac{a^{2} (1 + t^{4} + 2 t^{2})}{{(1 + t^{2})}^{2}} = \frac{a^{2} {(1 + t^{2})}^{2}}{{(1 + t^{2})}^{2}} = a^{2} \\ ∴ x^{2} + y^{2} = a^{2} w h i c h i s t h e e q u a t i o n o f a c i r c l e . \\ H e n c e, t h e g i v e n points l i e o n a c i r c l e . \end{array}$

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