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Exemplar Solution Class 11th

If p is the length of perpendicular from the origin on the line $\frac{x}{a} + \frac{y}{b}$ =1 and $a^{2}, p^{2}, b^{2}$ are in A.P., then show that $a^{2} + b^{2}$ =0.

3 Views|Posted a year ago

Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

Sol:

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

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Exemplar Solution Class 11th

If the lines $3 x - 4 y + 4 = 0$ and $6 x - 8 y - 7 = 0$ are tangents to a circle, then find the radius of the circle.
[Hint: Distance between given parallel lines gives the diameter of the circle.]

7 Views|Posted a year ago

Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} G i v e n e q u a t i o n a r e 3 x - 4 y + 4 = 0 \\ a n d 6 x - 8 y - 7 = 0 \Rightarrow 3 x - 4 y - \frac{7}{2} = 0 \\ Since \frac{3}{6} = \frac{- 4}{- 8} = \frac{1}{2} t h e n t h e l i n e s a r e p a r a l l e l . \\ S o, t h e distance b e t w e e n t h e p a r a l l e l l i n e s = | \frac{c_{1} - c_{2}}{\sqrt[]{a^{2} + b^{2}}} | = | \frac{4 + \frac{7}{2}}{\sqrt[]{{(3)}^{2} + {(- 4)}^{2}}} | \\ = | \frac{\frac{15}{2}}{5} | = \frac{3}{2} \\ D i a m e t e r = \frac{3}{2} \\ ∴ R a d i u s = \frac{3}{4} \\ H e n c e, t h e r e q u i r e d r a d i u s = \frac{3}{4} . \end{array}$

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Exemplar Solution Class 11th

P1, P2 are points on either of the two lines y – 3x = 2 at a distance of 5 units from their point of intersection. Find the coordinates of the foot of perpendiculars drawn from P?, P? on the bisector of the angle between the given lines.

(Hint: Lines are y = $\sqrt[]{3}$ x + 2 and y = – $\sqrt[]{3}$ x + 2 according as x ≥ 0 or x < 0

Y-axis is the bisector of the angles between the lines. P?, P? are the points on these lines at a distance of 5 units from the point of intersection of these lines which have a point on y-axis as the common foot of perpendiculars from these points. The y-coordinate of the foot of the perpendicular is given by 2 + 5 cos30°.)

7 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} G i v e n l i n e s a r e y - \sqrt[]{3} | x | = 2 \\ \Rightarrow y - \sqrt[]{3} x = 2, i f x \geq 0 \dots (i) \\ a n d y + \sqrt[]{3} x = 2, i f x < 0 \dots (i i) \\ S l o p e o f e q n . (i) i s t a n θ = \sqrt[]{3} ∴ θ = 6 0^{0} \\ S l o p e o f e q n . (i i) i s t a n θ = - \sqrt[]{3} ∴ θ = 12 0^{0} \\ S o l v i n g e q n . (i) a n d e q n . (i i) w e g e t \\ y - \sqrt[]{3} x = 2 \\ \underline{y + \sqrt[]{3} x = 2} \\ 2 y = 4 \Rightarrow y = 2 \\ P u t t i n g t h e v a l u e o f y i n e q n . (i) w e g e t \\ x = 0 \\ ∴ o f o f l i n e (i) a n d (i i) i s Q (0, 2) \\ ∴ Q O = 2 \\ I n Δ P E Q, \\ c o s 3 0^{0} = \frac{P Q}{Q E} \\ \Rightarrow \frac{\sqrt[]{3}}{2} = \frac{P Q}{5} \Rightarrow P Q = \frac{5 \sqrt[]{3}}{2} \\ ∴ O P = O Q + P Q \\ = 2 + \frac{5 \sqrt[]{3}}{2} \\ H e n c e, t h e c o o r d i n a t e s o f t h e f o o t o f p e r p e n d i c u l a r = (0, 2 + \frac{5 \sqrt[]{3}}{2}) \end{array}$

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Exemplar Solution Class 11th

If the sum of the distances of a moving point in a plane from the axes is 1, then find the locus of the point.

(Hint: Given that x + y = 1, which gives four sides of a square.

7 Views|Posted a year ago

Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

Sol:

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

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Exemplar Solution Class 11th

Find the equations of the lines through the point of intersection of the lines x – y + 1 = 0 and 2x – 3y + 5 = 0 and whose distance from the point (3, 2) is √5.

5 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} G i v e n e q u a t i o n s a r e \\ x - y + 1 = 0 \dots (i) \\ a n d 2 x - 3 y + 5 = 0 \dots (i i) \\ S o l v i n g e q n . (i) a n d e q n . (i i) w e g e t \\ 2 x - 2 y + 2 = 0 \\ 2 x - 3 y + 5 = 0 \\ \underline{(-) (+) (-)} \\ y - 3 = 0 ∴ y = 3 \\ F r o m e q n . (i) w e h a v e \\ x - 3 + 1 = 0 \Rightarrow x = 2 \\ S o, (2, 3) i s t h e o f o f e q n . (i) a n d e q n . (i i) . \\ L e t m b e t h e s l o p e o f t h e r e q u i r e d l i n e \\ ∴ E q u a t i o n o f t h e l i n e i s \\ y - 3 = m (x - 2) \\ \Rightarrow y - 3 = m x - 2 m \\ \Rightarrow m x - y + 3 - 2 m = 0 \\ Since, t h e p e r p e n d i c u l a r f r o m (3, 2) t o t h e l i n e i s \frac{7}{5} t h e n \\ \frac{7}{5} = | \frac{m (3) - 2 + 3 - 2 m}{\sqrt[]{m^{2} + 1}} | \\ \Rightarrow \frac{49}{25} = \frac{{(3 m - 2 + 3 - 2 m)}^{2}}{m^{2} + 1} \\ \Rightarrow \frac{49}{25} = \frac{{(m + 1)}^{2}}{m^{2} + 1} \\ \Rightarrow 49 m^{2} + 49 = 25 m^{2} + 50 m + 25 \\ \Rightarrow 49 m^{2} - 25 m^{2} - 50 m + 49 - 25 = 0 \\ \Rightarrow 24 m^{2} - 50 m + 24 = 0 \\ \Rightarrow 12 m^{2} - 25 m + 12 = 0 \\ \Rightarrow 12 m^{2} - 16 m - 9 m + 12 = 0 \\ \Rightarrow 4 m (3 m - 4) - 3 (3 m - 4) = 0 \\ \Rightarrow (3 m - 4) (4 m - 3) = 0 \\ \Rightarrow 3 m - 4 = 0 a n d 4 m - 3 = 0 \\ \Rightarrow m = \frac{4}{3}, \frac{3}{4} \\ E q u a t i o n o f t h e l i n e t a k i n g m = \frac{4}{3} i s \\ y - 3 = \frac{4}{3} (x - 2) \\ \Rightarrow 3 y - 9 = 4 x - 8 \Rightarrow 4 x - 3 y + 1 = 0 \\ E q u a t i o n o f t h e l i n e t a k i n g m = \frac{3}{4} i s \\ y - 3 = \frac{3}{4} (x - 2) \\ \Rightarrow 4 y - 12 = 3 x - 6 \Rightarrow 3 x - 4 y + 6 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n s a r e 4 x - 3 y + 1 = 0 a n d 3 x - 4 y + 6 = 0 . \end{array}$

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Exemplar Solution Class 11th

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.

3 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

Sol:

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

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Exemplar Solution Class 11th

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).)

5 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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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Exemplar Solution Class 11th

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?

4 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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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Exemplar Solution Class 11th

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).

4 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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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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.)

5 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

Sol:

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

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If p is the length of perpendicular from the origin on the line $\frac{x}{a} + \frac{y}{b}$ =1 and $a^{2}, p^{2}, b^{2}$ are in A.P., then show that $a^{2} + b^{2}$ =0.

If the lines $3 x - 4 y + 4 = 0$ and $6 x - 8 y - 7 = 0$ are tangents to a circle, then find the radius of the circle.
[Hint: Distance between given parallel lines gives the diameter of the circle.]

If the sum of the distances of a moving point in a plane from the axes is 1, then find the locus of the point.

(Hint: Given that x + y = 1, which gives four sides of a square.

Find the equations of the lines through the point of intersection of the lines x – y + 1 = 0 and 2x – 3y + 5 = 0 and whose distance from the point (3, 2) is √5.

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.

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).)

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?

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.)