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New answer posted

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Exemplar Solution Class 11th

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.)

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

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

Sol:

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

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New answer posted

3 months ago

Exemplar Solution Class 11th

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).

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

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

Sol:

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

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New answer posted

3 months ago

Exemplar Solution Class 11th

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°.)

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

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

Sol:

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

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New answer posted

3 months ago

Exemplar Solution Class 11th

If the intercept of a line between the coordinate axes is divided by the point (–5, 4) in the ratio 1: 2, then find the equation of the line.

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

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

Sol:

$\begin{array}{l} L e t a a n d b b e t h e o n t h e g i v e n l i n e . \\ ∴ C o o r d i n a t e s o f A a n d B a r e (a, 0) a n d (0, b) r e s p e c t i v e l y \\ ∴ - 5 = \frac{1 * 0 + 2 * a}{1 + 2} \\ \Rightarrow 2 a = - 15 \Rightarrow a = \frac{- 15}{2} [\begin{array}{l} ? X = \frac{m_{1} x_{2} + m_{2} x_{1}}{m_{1} + m_{2}} \\ a n d Y = \frac{m_{1} y_{2} + m_{2} y_{1}}{m_{1} + m_{2}} \end{array}] \\ ∴ A = (\frac{- 15}{2}, 0) \\ a n d 4 = \frac{1 * b + 0 * 2}{1 + 2} \\ \Rightarrow 4 = \frac{b}{3} \Rightarrow b = 12 \\ ∴ B = (0, 12) \\ S o, t h e e q u a t i o n o f l i n e A B i s \\ y - y_{1} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} (x - x_{1}) \\ \Rightarrow y - 0 = \frac{12 - 0}{0 + \frac{15}{2}} (x + \frac{15}{2}) \\ \Rightarrow y = \frac{12 * 2}{15} (x + \frac{15}{2}) \\ \Rightarrow y = \frac{8}{5} (x + \frac{15}{2}) \\ \Rightarrow 5 y = 8 x + 60 \\ \Rightarrow 8 x - 5 y + 60 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s 8 x - 5 y + 60 = 0 . \end{array}$

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New answer posted

3 months ago

Exemplar Solution Class 11th

For what values of a and b the intercepts cut off on the coordinate axes by the line ax + by + 8 = 0 are equal in length but opposite in signs to those cut off by the line 2x – 3y + 6 = 0 on the axes.

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

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

Sol:

$\begin{array}{l} T h e g i v e n e q u a t i o n a r e a x + b y + 8 = 0 \dots (i) \\ a n d 2 x - 3 y + 6 = 0 \dots (i i) \\ F r o m e q n . (i) w e g e t, \\ a x + b y + 8 = 0 \Rightarrow a x + b y = - 8 \\ \Rightarrow \frac{a}{- 8} x + \frac{b}{- 8} y = 1 \Rightarrow \frac{x}{\frac{- 8}{a}} + \frac{y}{\frac{- 8}{b}} = 1 \\ S o, t h e o n t h e a x e s a r e \frac{- 8}{a} a n d \frac{- 8}{b} \\ F r o m e q n . (i i) w e g e t, \\ 2 x - 3 y + 6 = 0 \Rightarrow 2 x - 3 y = - 6 \\ \Rightarrow \frac{2 x}{- 6} - \frac{3 y}{- 6} = 1 \Rightarrow \frac{x}{- 3} + \frac{y}{2} = 1 \\ S o, t h e o n t h e a x e s a r e - 3 a n d 2 . \\ A c c o r d i n g t o t h e q u e s t i o n \\ \frac{- 8}{a} = + 3 \Rightarrow a = - \frac{8}{3} \\ a n d \frac{- 8}{b} = - 2 \Rightarrow b = + 4 \\ H e n c e, t h e r e q u i r e d v a l u e s o f a a n d b a r e - \frac{8}{3} a n d 4 . \end{array}$

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New answer posted

3 months ago

Exemplar Solution Class 11th

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 3 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

3 months ago

Exemplar Solution Class 11th

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

3 months ago

Exemplar Solution Class 11th

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 3 Views

Vishal Baghel

Contributor-Level 10

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

Sol:

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

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New answer posted

3 months ago

Exemplar Solution Class 11th

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

3 months ago

Exemplar Solution Class 11th

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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