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New answer posted

a year ago

Class 11th Exemplar Solution

Find the equation of the line which passes through the point (–4, 3), and the portion of the line intercepted between the axes is divided internally in the ratio 5: 3 by this point.

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Vishal Baghel

Contributor-Level 10

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

Sol:

$\begin{array}{l} L e t A B b e a l i n e t h r o u g h a (- 4, 3) a n d m e e t s x a x i s a t A (a, 0) a n d \\ y a x i s a t B (0, b) . \\ ∴ - 4 = \frac{5 * 0 + 3 a}{5 + 3} \\ \Rightarrow - 4 = \frac{3 a}{8} \Rightarrow a = \frac{- 32}{3} [\begin{array}{l} ? X = \frac{m_{1} x_{2} + m_{2} x_{1}}{m_{1} + m_{2}} \\ a n d Y = \frac{m_{1} y_{2} + m_{2} y_{1}}{m_{1} + m_{2}} \end{array}] \\ a n d 3 = \frac{5 * b + 3 * 0}{5 + 3} \\ \Rightarrow 3 = \frac{5 b}{8} \Rightarrow b = \frac{24}{5} \\ I n t e r c e p t f o r m o f t h e l i n e i s \\ \frac{x}{\frac{- 32}{3}} + \frac{y}{\frac{24}{5}} = 1 \\ \Rightarrow \frac{- 3 x}{32} + \frac{5 y}{24} = 1 \\ \Rightarrow - 9 x + 20 y = 96 \Rightarrow 9 x - 20 y + 96 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s 9 x - 20 y + 96 = 0 . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

A straight line moves so that the sum of the reciprocals of its intercepts made on axes is constant. Show that the line passes through a fixed point.

(Hint: 1/a+1/b=1, where a + b =” constant” =1/k (say). This implies that a + b = k. Line passes through the fixed point (k, k).)

0 Follower 5 Views

Vishal Baghel

Contributor-Level 10

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

Sol:

$\begin{array}{l} I n t e r c e p t s f o r m o f a s t r a i g h t l i n e i s \\ \frac{x}{a} + \frac{y}{b} = 1 \\ w h e r e a a n d b a r e t h e m a d e b y t h e l i n e o n t h e a x e s . \\ G i v e n t h a t : \frac{1}{a} + \frac{1}{b} = \frac{1}{k} (s a y) \\ \Rightarrow \frac{k}{a} + \frac{k}{b} = 1 \\ w h i c h s h o w s t h a t t h e l i n e i s t h r o u g h t h e f i x e d (k, k) . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

In what direction should a line be drawn through the point (1, 2) so that its point of intersection with the line x + y = 4 is at a distance √5 from the given point?

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

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

Sol:

$\begin{array}{l} L e t t h e g i v e n l i n e x + y = 4 a n d t h e r e q u i r e d l i n e' l' a t B (a, b) \\ S l o p e o f l i n e' l' i s g i v e n b y m = \frac{b - 2}{a - 1} = t a n θ \dots (i) \\ G i v e n t h a t A B = \frac{\sqrt[]{6}}{3} \\ S o, b y f o r m u l a f o r A (1, 2) a n d B (a, b) \\ \sqrt[]{{(a - 1)}^{2} + {(b - 2)}^{2}} = \frac{\sqrt[]{6}}{3} \\ O n s q u a r i n g b o t h t h e s i d e \\ a^{2} + 1 - 2 a + b^{2} + 4 - 4 b = \frac{6}{9} \\ \Rightarrow a^{2} + b^{2} - 2 a - 4 b + 5 = \frac{6}{9} \dots (i i) \\ B (a, b) a l s o s a t i f i e s t h e e q n . x + y = 4 \\ ∴ a + b = 4 \dots (i i i) \\ O n s o l v i n g (i i) a n d (i i i), w e g e t a = \frac{3 \sqrt[]{3} + 1}{2 \sqrt[]{3}}, b = \frac{5 \sqrt[]{3} - 1}{2 \sqrt[]{3}} \\ P u t t i n g t h e v a l u e s o f a a n d b i n e q n . (i), w e h a v e \\ t a n θ = \frac{\frac{5 \sqrt[]{3} - 1}{2 \sqrt[]{3}} - 2}{\frac{3 \sqrt[]{3} + 1}{2 \sqrt[]{3}}} = \frac{5 \sqrt[]{3} - 1 - 4 \sqrt[]{3}}{3 \sqrt[]{3} + 1 - 2 \sqrt[]{3}} = \frac{\sqrt[]{3} - 1}{\sqrt[]{3} + 1} \\ ∴ t a n θ = t a n 1 5^{0} \Rightarrow θ = 1 5^{0} \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

A variable line passes through a fixed-point P. The algebraic sum of the perpendiculars drawn from the points (2, 0), (0, 2), and (1, 1) on the line is zero. Find the coordinates of the point P.

(Hint: Let the slope of the line be m. Then the equation of the line passing through the fixed-point P (x?, y?) is y – y ? = m (x – x?). Taking the algebraic sum of perpendicular distances equal to zero, we get y – 1 = m (x – 1). Thus (x?, y?) is (1, 1).

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

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

Sol:

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

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New answer posted

a year ago

Class 11th Exemplar Solution

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 5 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

a year ago

Class 11th Exemplar Solution

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 question posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Nine

Sol:

$\begin{array}{l} G i v e n t h a t \underset{h \to 0}{l i m} \frac{\sqrt[]{x + h} - \sqrt[]{x}}{h} \\ = \underset{h \to 0}{l i m} \frac{\sqrt[]{x + h} - \sqrt[]{x}}{h} \times \frac{\sqrt[]{x + h} + \sqrt[]{x}}{\sqrt[]{x + h} + \sqrt[]{x}} [R a t i o n a l i z i n g t h e] \\ = \underset{h \to 0}{l i m} \frac{x + h - x}{h [\sqrt[]{x + h} + \sqrt[]{x}]} = \underset{h \to 0}{l i m} \frac{h}{h [\sqrt[]{x + h} + \sqrt[]{x}]} \\ = \underset{h \to 0}{l i m} \frac{1}{\sqrt[]{x + h} + \sqrt[]{x}} \\ T a k i n g, w e h a v e \\ \frac{1}{\sqrt[]{x} + \sqrt[]{x}} = \frac{1}{2 \sqrt[]{x}} . \end{array}$

0 Follower 2 Views

New answer posted

a year ago

Class 11th Exemplar Solution

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 5 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

a year ago

Class 11th Exemplar Solution

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 4 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

a year ago

Class 11th Exemplar Solution

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 5 Views

Vishal Baghel

Contributor-Level 10

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

Sol:

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

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