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New answer posted

a year ago

Class 11th Exemplar Solution

The tangent of angle between the lines whose intercepts on the axes are (a,-b) and (b,-a), respectively, is $^{}$

(a) $\frac{a^{2} - b^{2}}{a b}$

(b) $\frac{b^{2} - a^{2}}{2}$

(c) $\frac{b^{2} - a^{2}}{2 a b}$

(d) None of these

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

Class 11th Exemplar Solution

The equation of the line passing through the point (1, 2) and perpendicular to the line x+y+1=0 is

(a) Y-x+1=0

(b) Y-x-1=0

(c) Y-x+2=0

(d) Y-x-2=0

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

Class 11th Exemplar Solution

The equation of the straight line passing through the point (3, 2) and perpendicular to the line y=x is

(a) X-y=5

(b) X+y=5

(c) X+y=1

(d) X-y=1

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} E q u a t i o n o f l i n e' l' i s g i v e n b y \\ y - y_{1} = m (x - x_{1}) \\ l i s t h r o u g h t h e P (3, 2) . \\ ∴ y - 2 = m (x - 3) \\ \Rightarrow y = m x + 2 - 3 m \dots (i) \\ i t i s g i v e n t h a t l i n e s y = x a n d' l' a r e p e r p e n d i c u l a r t o e a c h o t h e r, \\ ∴ m * 1 = - 1 [? m_{1} * m_{2} = - 1] \\ m = - 1 \\ P u t m = - 1 i n e q n . (i), w e g e t \\ y = - x + 2 - 3 (- 1) \\ y = - x + 5 \\ x + y = 5 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

Slope of a line which cuts off intercepts of equal lengths on the axes is

(a) -1

(b) -0

(c) 2

(d) 3

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} f o r m o f a l i n e i s \\ \frac{x}{a} + \frac{y}{b} = 1 \\ \Rightarrow \frac{x}{a} + \frac{y}{a} = 1 [? a = b] \\ \Rightarrow x + y = a \\ \Rightarrow y = - x + a \\ ∴ S l o p e i s - 1 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

A line cutting off intercept –3 from the y-axis and the tangent at angle θ to the x-axis is tanθ = $\frac{3}{5}$ , its equation is

(a) 5y-3x+15=0

(b) 3y-5x+15=0

(c) 5y-3x-15=0

(d) None of these

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} t h e l i n e s c u t o f f - 3 o n y a x i s t h e n t h e l i n e i s t h r o u g h t h e (0, - 3) . \\ G i v e n t h a t : t a n θ = \frac{3}{5} \Rightarrow S l o p e o f t h e l i n e m = \frac{3}{5} \\ S o, t h e e q u a t i o n o f t h e l i n e i s \\ y - y_{1} = m (x - x_{1}) \\ \Rightarrow y + 3 = \frac{3}{5} (x - 0) \\ \Rightarrow 5 y + 15 = 3 x \\ \Rightarrow 3 x - 5 y - 15 = 0 \Rightarrow 5 y - 3 x + 15 = 0 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

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.

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

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

Sol:

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

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New answer posted

a year ago

Class 11th Exemplar Solution

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.]

0 Follower 7 Views

Payal Gupta

Contributor-Level 10

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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New answer posted

a year ago

Class 11th Exemplar Solution

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 b

...more

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

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0 Follower 7 Views

Vishal Baghel

Contributor-Level 10

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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New answer posted

a year ago

Class 11th Exemplar Solution

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.

0 Follower 7 Views

Vishal Baghel

Contributor-Level 10

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

Sol:

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

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New answer posted

a year ago

Class 11th Exemplar Solution

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.

0 Follower 5 Views

Vishal Baghel

Contributor-Level 10

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

Sol:

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

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