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New answer posted

a year ago

Class 11th Exemplar Solution

The line joining the points (3, –4) and (–2, 6) is perpendicular to the line joining the points (–3, 6) and (9, –18).

0 Follower 5 Views

Vishal Baghel

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} T h e g i v e n a r e (3, - 4) a n d (- 2, 6), (- 3, 6) a n d (9, - 18) . \\ S l o p e o f t h e l i n e j o i n i n g t h e (3, - 4) a n d (- 2, 6) \\ m_{1} = \frac{6 + 4}{- 2 - 3} = \frac{10}{- 5} = - 2 \\ S l o p e o f t h e l i n e j o i n i n g t h e (- 3, 6) a n d (9, - 18) \\ m_{2} = \frac{- 18 - 6}{9 + 3} = \frac{- 24}{12} = - 2 \\ S i n c e m_{1} = m_{2} = - 2 \\ S o, t h e l i n e s a r e p a r a l l e l a n d n o t p e r p e n d i c u l a r . \\ H e n c e t h e g i v e n s t a t e m e n t i s F a l s e . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

The lines $a x + 2 y + 1 = 0, b x + 3 y + 1 = 0,$ and $c x + 4 y + 1 = 0$ are concurrent if a, b, c are in G.P.

0 Follower 6 Views

Vishal Baghel

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

Class 11th Exemplar Solution

The line $\frac{x}{a} + \frac{y}{b} = 1$ moves in such a way that $\frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{c^{2}}$ , where c is a constant. The locus of the foot of the perpendicular from the origin on the given line is $x^{2} + y^{2} = c^{2}$ .

0 Follower 5 Views

Vishal Baghel

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

Class 11th Exemplar Solution

The equation of the line joining the point (3, 5) to the point of intersection of the lines 4x + y – 1 = 0 and 7x – 3y – 35 = 0 is equidistant from the points (0, 0) and (8, 34).

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n e q u a t i o n s a r e \\ 4 x + y - 1 = 0 \dots (i) \\ a n d 7 x - 3 y - 35 = 0 \dots (i i) \\ F r o m e q n . (i) y = 1 - 4 x \dots (i i i) \\ P u t t i n g t h e v a l u e o f y i n e q n . (i i) w e g e t \\ 7 x - 3 (1 - 4 x) - 35 = 0 \\ \Rightarrow 7 x - 3 + 12 x - 35 = 0 \\ \Rightarrow 19 x - 38 = 0 \\ \Rightarrow x = 2 \\ F r o m e q n . (i i i) y = 1 - 4 * 2 \Rightarrow y = - 7 \\ T h e o f i s (2, - 7) . \\ E q u a t i o n o f l i n e j o i n i n g t h e (3, 5) t o t h e (2, - 7) i s \\ y - 5 = \frac{- 7 - 5}{2 - 3} (x - 3) \\ \Rightarrow y - 5 = 12 (x - 3) \\ \Rightarrow y - 5 = 12 x - 36 \\ \Rightarrow 12 x - y - 31 = 0 \dots (i v) \\ o f e q n . (i v) f r o m t h e (0, 0) = | \frac{- 31}{\sqrt[]{{(12)}^{2} + {(- 1)}^{2}}} | = \frac{31}{\sqrt[]{145}} \\ o f e q n . (i v) f r o m t h e (8, 34) = | \frac{12 * 8 - 34 - 31}{\sqrt[]{{(12)}^{2} + {(- 1)}^{2}}} | = | \frac{96 - 65}{\sqrt[]{145}} | = \frac{31}{\sqrt[]{145}} \\ H e n c e t h e g i v e n s t a t e m e n t i s T r u e . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

The vertex of an equilateral triangle is (2, 3) and the equation of the opposite side is x + y = 2. Then the other two sides are $y – 3 = (2 \pm \sqrt[]{3}) (x – 2)$ .

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

Class 11th Exemplar Solution

The straight line 5x + 4y = 0 passes through the point of intersection of the straight lines x + 2y – 10 = 0 and 2x + y + 5 = 0.

0 Follower 5 Views

Vishal Baghel

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n e q u a t i o n s a r e x + 2 y - 10 = 0 \dots (i) \\ a n d 2 x + y + 5 = 0 \dots (i i) \\ F r o m e q n . (i) x = 10 - 2 y \dots (i i i) \\ P u t t i n g t h e v a l u e o f x i n e q n . (i i) w e g e t \\ 2 (10 - 2 y) + y + 5 = 0 \\ \Rightarrow 20 - 4 y + y + 5 = 0 \\ \Rightarrow - 3 y + 25 = 0 \\ \Rightarrow y = \frac{25}{3} \\ P u t t i n g t h e v a l u e o f y i n e q n . (i i i) w e g e t \\ x = 10 - 2 (\frac{25}{3}) \\ = \frac{30 - 50}{3} = \frac{- 20}{3} \\ ∴ = (\frac{- 20}{3}, \frac{25}{3}) \\ I f t h e g i v e n l i n e 5 x + 4 y = 0 t h r o u g h t h e (\frac{- 20}{3}, \frac{25}{3}) t h e n \\ 5 (\frac{- 20}{3}) + 4 (\frac{25}{3}) = 0 \\ \Rightarrow \frac{- 100}{3} + \frac{100}{3} = 0 \\ \Rightarrow 0 = 0 s a t i s f i e d . \\ S o, t h e g i v e n l i n e t h r o u g h t h e o f o f t h e g i v e n l i n e s . \\ H e n c e, t h e g i v e n s t a t e m e n t i s T r u e . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

Equation of the line passing through the point $(a {c o s}^{3} θ, a {s i n}^{3} θ)$ and perpendicular to the line $x s e c θ + y c o s e c θ = a$ is $x c o s θ – y s i n θ = a s i n 2 θ$ .

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} E q u a t i o n o f a n y l i n e p e r p e n d i c u l a r t o x s e c θ + y c o s e c θ = a i s \\ x c o s e c θ - y s e c θ = k \dots (i) \\ I f e q n . (i) p a s s e s t h r o u g h (a c o s^{3} θ, a s i n^{3} θ) t h e n \\ a c o s^{3} θ . c o s e c θ - a s i n^{3} θ . s e c θ = k \\ \Rightarrow \frac{a c o s^{3} θ}{s i n θ} - \frac{a s i n^{3} θ}{c o s θ} = k \\ ∴ e q u a t i o n i s \\ x c o s e c θ - y s e c θ = \frac{a c o s^{3} θ}{s i n θ} - \frac{a s i n^{3} θ}{c o s θ} \\ \Rightarrow \frac{x}{s i n θ} - \frac{y}{c o s θ} = a [\frac{c o s^{4} θ - s i n^{4} θ}{s i n θ c o s θ}] \\ \Rightarrow \frac{x c o s θ - y s i n θ}{s i n θ c o s θ} = a [\frac{(c o s^{2} θ + s i n^{2} θ) (c o s^{2} θ - s i n^{2} θ)}{s i n θ c o s θ}] \\ \Rightarrow x c o s θ - y s i n θ = a (c o s^{2} θ - s i n^{2} θ) \\ \Rightarrow x c o s θ - y s i n θ = a c o s 2 θ \\ S o, t h e g i v e n s t a t e m e n t i s F a l s e . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

The points A(–2, 1), B(0, 5), C(–1, 2) are collinear.

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n a r e A (- 2, 1), B (0, 5), C (- 1, 2) \\ A r e a o f Δ A B C = \frac{1}{2} | \begin{matrix} - 2 & 1 & 1 \\ 0 & 5 & 1 \\ - 1 & 2 & 1 \end{matrix} | \\ = \frac{1}{2} | - 2 | \begin{matrix} 5 & 1 \\ 2 & 1 \end{matrix} | - 1 | \begin{matrix} 0 & 1 \\ - 1 & 1 \end{matrix} | + 1 | \begin{matrix} 0 & 5 \\ - 1 & 2 \end{matrix} | | \\ = \frac{1}{2} | - 2 (5 - 2) - 1 (0 + 1) + 1 (0 + 5) | \\ = \frac{1}{2} | - 2 * 3 - 1 * 1 + 1 * 5 | = \frac{1}{2} | - 6 - 1 + 5 | = \frac{1}{2} | - 2 | = 1 s q . u n i t \\ S o, t h e g i v e n a r e n o t c o l l i n e a r . \\ S o, t h e g i v e n s t a t e m e n t i s F a l s e . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

If the vertices of a triangle have integral coordinates, then the triangle cannot be equilateral.

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e k n o w t h a t i f t h e v e r t i c e s o f t r i a n g l e h a s c o o r d i n a t e s, t h e n t h e t r i a n g l e c a n n o t b e \\ e q u i l a t e r a l . \\ S o, t h e g i v e n s t a t e m e n t i s T r u e . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

Locus of the midpoints of the portion of the line $x s i n θ + y c o s θ = p$ intercepted between the axes is ____.

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

This is a Fill in the Blank Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n e q u a t i o n o f t h e l i n e i s \\ x c o s θ + y s i n θ = p \dots (i) \\ L e t C (h, k) b e t h e m i d o f t h e g i v e n l i n e A B w h e r e i t m e e t s t h e t w o a x i s a t \\ A (a, 0) a n d B (0, b) . \\ (a, 0) l i e s o n e q n . (i) t h e n \\ a c o s θ + 0 = p \\ \Rightarrow a = \frac{p}{c o s θ} \dots (i i) \\ B (0, b) a l s o l i e s o n e q n . (i) t h e n \\ 0 + b s i n θ = p \\ \Rightarrow b = \frac{p}{s i n θ} \dots (i i i) \\ C (h, k) i s t h e m i d o f A B \\ ∴ h = \frac{0 + a}{2} \Rightarrow a = 2 h \\ a n d k = \frac{b + 0}{2} \Rightarrow b = 2 k \\ P u t t i n g t h e v a l u e s o f a a n d b i s e q n . (i i) a n d (i i i) w e g e t \\ 2 h = \frac{p}{c o s θ} \Rightarrow c o s θ = \frac{p}{2 h} \dots (i v) \\ a n d 2 k = \frac{p}{s i n θ} \Rightarrow s i n θ = \frac{p}{2 k} \dots (v) \\ S q u a r i n g a n d a d d i n g e q n . (i v) a n d (v) w e g e t \\ c o s^{2} θ + s i n^{2} θ = \frac{p^{2}}{4 h^{2}} + \frac{p^{2}}{4 k^{2}} \\ \Rightarrow 1 = \frac{p^{2}}{4 h^{2}} + \frac{p^{2}}{4 k^{2}} \\ S o, t h e l o c u s o f t h e m i d i s \\ 1 = \frac{p^{2}}{4 x^{2}} + \frac{p^{2}}{4 y^{2}} \\ \Rightarrow 4 x^{2} y^{2} = p^{2} (x^{2} + y^{2}) \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 4 x^{2} y^{2} = p^{2} (x^{2} + y^{2}) . \end{array}$

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