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Exemplar Solution Class 11th

The distance between the lines $y = m x + c_{1}$ and $y = m x + c_{2}$ is

(a) $\frac{c_{1} - c_{2}}{\sqrt[]{1 + m^{2}}}$

(b) $\frac{∣ c_{1} - c_{2} ∣}{\sqrt[]{1 + m^{2}}}$

(c) $\frac{2 (c_{1} - c_{2})}{\sqrt[]{1 + m^{2}}}$

(d) 0

4 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 Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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Exemplar Solution Class 11th

The equations of the lines passing through the point (1, 0) and at a distance $\frac{\sqrt[]{3}}{2}$ from the origin, are

(a) $3 x + y – 3 = 0, 3 x – y – 3 = 0$ (b)  $3 x + y + 3 = 0, 3 x – y + 3 = 0$ (c) $x + 3 y – 3 = 0, x – 3 y – 3 = 0$  (d)  None of these

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 Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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Exemplar Solution Class 11th

The equations of the lines which pass through the point (3, –2) and are inclined at 60° to the line 3x + y = 1 is

(a) $y + 2 = 0, 3 x – y – 2 - 3 \sqrt[]{3} = 0$ (b) $x – 2 = 0, 3 x – y + 2 + 3 \sqrt[]{3} = 0$ 

(c) $3 x – y – 2 - 3 \sqrt[]{3} = 0$  (d) None of these

5 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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Exemplar Solution Class 11th

The distance of the point of intersection of the lines 2x – 3y + 5 = 0 and 3x + 4y = 0 from the line 5x – 2y = 0 is

(a) $\frac{130}{17}$ 

(b) $\frac{13 \sqrt[]{7}}{29}$ 

(c) $\frac{130}{7}$ (d) None of these

4 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n e q u a t i o n s a r e : \\ 2 x - 3 y + 5 = 0 \dots (i) \\ 3 x + 4 y = 0 \dots (i i) \\ F r o m e q n . (i i) w e g e t, \\ 4 y = - 3 x \Rightarrow y = \frac{- 3}{4} x \dots (i i i) \\ P u t t i n g t h e v a l u e o f y i n e q n . (i) w e h a v e \\ 2 x - 3 (\frac{- 3}{4} x) + 5 = 0 \\ \Rightarrow 8 x + 9 x + 20 = 0 \\ \Rightarrow 17 x + 20 = 0 \Rightarrow x = \frac{- 20}{17} \\ P u t t i n g t h e v a l u e o f x i n e q n . (i i i) w e g e t \\ y = \frac{- 3}{4} (\frac{- 20}{17}) = \frac{15}{17} \\ ∴ o f i s (\frac{- 20}{17}, \frac{15}{17}) \\ N o w p e r p e n d i c u l a r f r o m t h e (\frac{- 20}{17}, \frac{15}{17}) t o t h e g i v e n l i n e 5 x - 2 y = 0 i s \\ | \frac{5 (\frac{- 20}{17}) - 2 (\frac{15}{17})}{\sqrt[]{25 + 4}} | = | \frac{\frac{- 100}{17} - \frac{30}{17}}{\sqrt[]{29}} | = \frac{130}{17 \sqrt[]{29}} \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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Exemplar Solution Class 11th

If the line $\frac{x}{a} + \frac{y}{b} = 1$ passes through the points (2, –3) and (4, –5), then (a, b) is

(a) (1, 1)

(b) (–1, 1)

(c) (1, –1)

(d) (–1, –1)

3 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} E q u a t i o n o f l i n e t h r o u g h t h e (2, - 3) a n d (4, - 5) i s \\ y + 3 = \frac{- 5 + 3}{4 - 2} (x - 2) \\ \Rightarrow y + 3 = \frac{- 2}{2} (x - 2) \\ \Rightarrow y + 3 = - (x - 2) \\ \Rightarrow y + 3 = - x + 2 \\ \Rightarrow x + y = - 1 \\ \Rightarrow \frac{x}{- 1} + \frac{y}{- 1} = 1 (I n t e r c e p t f o r m) \\ ∴ a = - 1, b = - 1 \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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Exemplar Solution Class 11th

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

3 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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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Exemplar Solution Class 11th

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

4 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 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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Exemplar Solution Class 11th

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

5 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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Exemplar Solution Class 11th

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

(a) -1

(b) -0

(c) 2

(d) 3

4 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 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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Exemplar Solution Class 11th

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

4 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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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The distance between the lines $y = m x + c_{1}$ and $y = m x + c_{2}$ is

(a) $\frac{c_{1} - c_{2}}{\sqrt[]{1 + m^{2}}}$

(b) $\frac{∣ c_{1} - c_{2} ∣}{\sqrt[]{1 + m^{2}}}$

(c) $\frac{2 (c_{1} - c_{2})}{\sqrt[]{1 + m^{2}}}$

(d) 0

If the line $\frac{x}{a} + \frac{y}{b} = 1$ passes through the points (2, –3) and (4, –5), then (a, b) is

(a) (1, 1)

(b) (–1, 1)

(c) (1, –1)

(d) (–1, –1)

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

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

(a) -1

(b) -0

(c) 2

(d) 3