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Exemplar Solution Class 11th

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.

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Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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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Exemplar Solution Class 11th

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 $x c o s θ – y s i n θ = a s i n 2 θ$ .

3 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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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Exemplar Solution Class 11th

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

4 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

Sol:

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

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Exemplar Solution Class 11th

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

3 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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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Exemplar Solution Class 11th

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 ____.

3 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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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Exemplar Solution Class 11th

A point moves so that the square of its distance from the point (3, –2) is numerically equal to its distance from the line 5x – 12y = 3. The equation of its locus is ____.

3 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} T h e g i v e n e q u a t i o n o f l i n e i s 5 x - 12 y = 3 a n d t h e g i v e n i s (3, - 2) . \\ L e t (a, b) b e a n y m o v i n g \\ ∴ b e t w e e n (a, b) a n d t h e (3, - 2) = \sqrt[]{{(a - 3)}^{2} + {(b + 2)}^{2}} \\ a n d t h e o f (a, b) f r o m t h e l i n e 5 x - 12 y = 3 \\ = | \frac{5 a - 12 b - 3}{\sqrt[]{25 + 144}} | = | \frac{5 a - 12 b - 3}{13} | \\ A c c o r d i n g t o t h e q u e s t i o n, w e h a v e \\ {[\sqrt[]{{(a - 3)}^{2} + {(b + 2)}^{2}}]}^{2} = | \frac{5 a - 12 b - 3}{13} | \\ T a k i n g n u m e r i c a l v a l u e s, w e h a v e \\ {(a - 3)}^{2} + {(b + 2)}^{2} = \frac{5 a - 12 b - 3}{13} \\ \Rightarrow a^{2} - 6 a + 9 + b^{2} + 4 b + 4 = \frac{5 a - 12 b - 3}{13} \\ \Rightarrow a^{2} + b^{2} - 6 a + 4 b + 13 = \frac{5 a - 12 b - 3}{13} \\ \Rightarrow 13 a^{2} + 13 b^{2} - 78 a + 52 b + 169 = 5 a - 12 b - 3 \\ \Rightarrow 13 a^{2} + 13 b^{2} - 83 a + 64 b + 172 = 0 \\ S o, t h e l o c u s o f t h e i s 13 a^{2} + 13 b^{2} - 83 a + 64 b + 172 = 0 . \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 13 a^{2} + 13 b^{2} - 83 a + 64 b + 172 = 0 . \end{array}$

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Exemplar Solution Class 11th

The points (3, 4) and (2, –6) are situated on the ____ of the line 3x – 4y – 8 = 0.

3 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} G i v e n l i n e i s 3 x - 4 y - 8 = 0 \dots (i) \\ a n d g i v e n t h e a r e (3, 4) a n d (2, - 6) . \\ F o r (3, 4), l i n e b e c o m e s = 3 (3) - 4 (4) - 8 \\ = 9 - 16 - 8 = 9 - 24 = - 15 < 0 \\ F o r (2, - 6), l i n e b e c o m e s = 3 (2) - 4 (- 6) - 8 \\ = 6 + 24 - 8 = 30 - 8 = 22 > 0 \\ S o, t h e (3, 4) a n d (2, - 6) a r e s i t u a t e d o n t h e o p p o s i t e s i d e s o f 3 x - 4 y - 8 = 0 . \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s o p p o s i t e . \end{array}$

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Exemplar Solution Class 11th

Equations of the lines through the point (3, 2) and making an angle of 45° with the line x – 2y = 3 are ____.

5 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} G i v e n l i n e i s x - 2 y = 3 a n d t h e i s (3, 2) \\ E q u a t i o n o f a l i n e t h r o u g h t h e (3, 2) i s \\ A n g l e b e t w e e n e q n . (i) a n d t h e g i v e n l i n e x - 2 y = 3 w h o s e s l o p e i s \frac{1}{2} \\ ∴ t a n θ = | \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} | \\ \Rightarrow t a n 4 5^{0} = | \frac{m - \frac{1}{2}}{1 + m \times \frac{1}{2}} | \\ \Rightarrow 1 = | \frac{m - \frac{1}{2}}{1 + \frac{m}{2}} | \Rightarrow \frac{m - \frac{1}{2}}{1 + \frac{m}{2}} = \pm 1 \\ T a k i n g (+) s i g n \\ \Rightarrow \frac{m - \frac{1}{2}}{1 + \frac{m}{2}} = 1 \Rightarrow m - \frac{1}{2} = 1 + \frac{m}{2} \\ \Rightarrow m - \frac{m}{2} = 1 + \frac{1}{2} \Rightarrow \frac{m}{2} = \frac{3}{2} \Rightarrow m = 3 \\ T a k i n g (-) s i g n \\ \Rightarrow \frac{m - \frac{1}{2}}{1 + \frac{m}{2}} = - 1 \Rightarrow m - \frac{1}{2} = - 1 - \frac{m}{2} \\ \Rightarrow m + \frac{m}{2} = - 1 + \frac{1}{2} \Rightarrow \frac{3 m}{2} = \frac{1}{2} \Rightarrow m = - \frac{1}{3} \\ S o, t h e r e q u i r e d e q u a t i o n s a r e, \\ W h e n m = 3, \\ y - 2 = 3 (x - 3) \\ \Rightarrow y - 2 = 3 x - 9 \\ \Rightarrow 3 x - y - 7 = 0 \\ W h e n m = - \frac{1}{3}, \\ y - 2 = - \frac{1}{3} (x - 3) \\ \Rightarrow 3 y - 6 = - x + 3 \\ \Rightarrow x + 3 y - 9 = 0 \\ H e n c e, t h e v a l u e o f t h e f i l l e r a r e 3 x - y - 7 = 0 a n d x + 3 y - 9 = 0 . \end{array}$

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Exemplar Solution Class 11th

The line which cuts off equal intercepts from the axes and passes through the point (1, –2) is ____.

3 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

Sol:

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

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Exemplar Solution Class 11th

If a, b, c are in A.P., then the straight lines ax + by + c = 0 will always pass through ____.

3 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

Sol:

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

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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.

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 $x c o s θ – y s i n θ = a s i n 2 θ$ .

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

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

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 ____.

A point moves so that the square of its distance from the point (3, –2) is numerically equal to its distance from the line 5x – 12y = 3. The equation of its locus is ____.

The points (3, 4) and (2, –6) are situated on the ____ of the line 3x – 4y – 8 = 0.

Equations of the lines through the point (3, 2) and making an angle of 45° with the line x – 2y = 3 are ____.

The line which cuts off equal intercepts from the axes and passes through the point (1, –2) is ____.

If a, b, c are in A.P., then the straight lines ax + by + c = 0 will always pass through ____.