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New answer posted

a year ago

Class 11th Exemplar Solution

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 ____.
0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

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

Sol:

The  given  equation  of  line  is  5x−12y=3  and  the  given    is  (3,−2).Let  (a,b)  be  any  moving  ∴    between  (a,b)  and  the    (3,−2)=(a−3)2+(b+2)2and  the    of  (a,b)  from  the  line  5x−12y=3                    =|5a−12b−325+144|=|5a−12b−313|According  to  the  question,  we  have     [(a−3)2+(b+2)2]2=|5a−12b−313|Taking  numerical  values,  we  have               (a−3)2+(b+2)2=5a−12b−313⇒a2−6a+9+b2+4b+4=5a−12b−313⇒       a2+b2−6a+4b+13=5a−12b−313⇒13a2+13b2−78a+52b+169=5a−12b−3⇒13a2+13b2−83a+64b+172=0So,  the  locus  of  the    is  13a2+13b2−83a+64b+172=0.Hence,  the  value  of  the  filler  is  13a2+13b2−83a+64b+172=0.

New answer posted

a year ago

Class 11th Exemplar Solution

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

V
Vishal Baghel

Contributor-Level 10

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

Sol:

Given  line  is  3x−4y−8=0                                       …(i)and  given  the    are  (3,4)  and  (2,−6).For    (3,4),  line  becomes=3(3)−4(4)−8              =9−16−8=9−24=−15<0For    (2,−6),  line  becomes=3(2)−4(−6)−8              =6+24−8=30−8=22>0So,  the    (3,4)  and  (2,−6)  are  situated  on  the  opposite  sides  of  3x−4y−8=0.Hence,  the  value  of  the  filler  is  opposite.

New answer posted

a year ago

Class 11th Exemplar Solution

Equations of the lines through the point (3, 2) and making an angle of 45° with the line x – 2y = 3 are ____.
0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

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

Sol:

Given  line  is  x−2y=3  and  the    is  (3,2)Equation  of  a  line    through  the    (3,2)  isAngle  between  eqn.(i)  and  the  given  line  x−2y=3  whose  slope  is  12∴             tanθ=|m1−m21+m1m2|⇒      tan450=|m−121+m*12|⇒                   1=|m−121+m2|        ⇒m−121+m2=±1Taking  (+)  sign⇒        m−121+m2=1         ⇒m−12=1+m2⇒        m−m2=1+12  ⇒m2=32    ⇒m=3Taking  (−)  sign⇒        m−121+m2=−1         ⇒m−12=−1−m2⇒        m+m2=−1+12  ⇒3m2=12    ⇒m=−13So,  the  required  equations  are,When  m=3,               y−2=3(x−3)⇒           y−2=3x−9⇒3x−y−7=0When  m=−13,          y−2=−13(x−3)⇒    3y−6=−x+3⇒x+3y−9=0Hence,  the  value  of  the  filler  are  3x−y−7=0  and  x+3y−9=0.

New answer posted

a year ago

Class 11th Exemplar Solution

The line which cuts off equal intercepts from the axes and passes through the point (1, –2) is ____.
0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

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

Sol:

Intercept  form  of  the  line  is  xa+yb=1                                      …(i)Given  that  a=b∴            xa+ya=1         ⇒x+y=a                                                     …(ii)If  the  line  (i)  passes  through  (1,−2)  we  get                 1−2=a         ⇒a=−1So,  the  required  equation  is  x+y=−1       ⇒x+y+1=0Hence,  the  value  of  the  filler  is  x+y+1=0.

New answer posted

a year ago

Class 11th Exemplar Solution

If a, b, c are in A.P., then the straight lines ax + by + c = 0 will always pass through ____.
0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

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

Sol:

Given  equation  is  ax+by+c=0                           …(i)  a,b  and  c  are  in  A.P.∴                     b=a+c2⇒           a+c=2b⇒a−2b+c=0                                                              …(ii)Comparing  eqn.(i)  with  eqn.(ii)  we  get,        x=1,  y=−2So,  the  line  will  pass  through  (1,−2)Hence,  the  value  of  the  filler  is  (1,−2).

New answer posted

a year ago

Class 11th Exemplar Solution

One vertex of the equilateral triangle with centroid at the origin and one side as x + y – 2 = 0 is

(a) (–1, –1)

(b) (2, 2)

(c)  (–2, –2)

(d) (2, –2)
0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

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     ( x 1 , y 1 ) . A D ⊥ B C     a n d     l e t     ( a , b )     b e     t h e     c o o r d i n a t e s     o f     D . G i v e n     t h a t     t h e     c e n t r o i d     G     l i e s     a t     t h e     o r i g i n     i . e . , ( 0 , 0 ) Since,  the  centroid  of  a  triangle,  divides  the  median  in  the  ratio  1:2 S o ,                         0 = 1 * x 1 + 2 * a 1 + 2                     ⇒ x 1 + 2 a = 0                                                                                           … ( i ) a n d                       0 = 1 * y 1 + 2 * b 1 + 2                   ⇒ y 1 + 2 b = 0                                                                                           … ( i i ) E q u a t i o n s     o f     B C     i s     g i v e n     b y     x + y − 2 = 0                                                                                       … ( i i i ) Point  D(a,b)  lies  on  the  line  x+y−2=0 S o ,                                                                                                                     a + b − 2 = 0                                                                                           … ( i v ) S l o p e     o f     e q n . ( i i i )     i s     = − 1 a n d     t h e     S l o p e     o f     A G = y 1 − 0 x 1 − 0 = y 1 x 1 Since,  they  are  perpendicular  to  each  other ∴                       − 1 * y 1 x 1 = − 1       ⇒ y 1 = x 1 F r o m     e q n . ( i )     a n d     ( i i )     w e     g e t                               x 1 + 2 a = 0         ⇒ 2 a = − x 1                               y 1 + 2 b = 0         ⇒ 2 b = − y 1 F r o m     e q n . ( i v )     w e     g e t                         a + b − 2 = 0 ⇒               a + a − 2 = 0 ⇒                         2 a − 2 = 0             ⇒ a = 1     a n d     b = 1               [ ? a = b ] ∴               x 1 = − 2 * 1 = − 2 a n d     y 1 = − 2 * 1 = − 2 H e n c e ,     t h e     c o r r e c t     o p t i o n     i s     ( c ) .

 

New answer posted

a year ago

Class 11th Exemplar Solution

The ratio in which the line 3x + 4y + 2 = 0 divides the distance between the lines 3x + 4y + 5 = 0 and 3x + 4y – 5 = 0 is

(a) 1 : 2

(b) 3 : 7

(c) 2 : 3

(d)  2 : 5
0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

Given  equations  are                   3x+4y+5=0                                        …(i)                   3x+4y−5=0                                       …(ii)and          3x+4y+2=0                                       …(iii)Clearly,  eqn.(i),  (ii)  and  (iii)  are  parallel  to  each  other  as  the  coefficients  of  x  and  y  are  same.  between  parallel  lines  (i)  and  (iii)  we  get            =|5−2(3)2+(4)2|=35         [?  between  the  two  parallel  lines=|c1−c2|a2+b2]  between  parallel  lines  (ii)  and  (iii)  we  get            =|−5−2(3)2+(4)2|=75∴  Ratio  between  the  35:75=3:7Hence,  the  correct  option  is  (b).

New answer posted

a year ago

Class 11th Exemplar Solution

A line passes through (2, 2) and is perpendicular to the line 3x + y = 3. Its y-intercept is

(a) 1 3

(b) 2 3   

(c) 1

(d)   4 3  

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

Any  line  perpendicular  to  3x+y=3            x−3y=λ         (λ=)If  it  passes  through  the    (2,2)  then        2−3(2)=λ      ⇒λ=−4∴    equation  is  x−3y=−4⇒       −3y=−x−4⇒             y=13x+43               [?y=mx+c]So,  the  y  is  43.Hence,  the  correct  option  is  (d).

New answer posted

a year ago

Class 11th Exemplar Solution

A point equidistant from the lines 4x + 3y + 10 = 0, 5x – 12y + 26 = 0, and 7x + 24y – 50 = 0 is

(a) (1, –1)

(b) (1, 1)

(c) (0, 0)

(d) (0, 1)
0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

Given  equations  are                        4x+3y+10=0                                       …(i)                     5x−12y+26=0                                       …(ii)and             7x+24y−50=0                                      …(iii)Let  (x1,y1)  be  any      from  eqn.(i),  eqn.(ii)  and  eqn.(iii)  of  (x1,y1)  from  eqn.(i)            =|4x1+3y1+1016+9|=|4x1+3y1+105|  of  (x1,y1)  from  eqn.(ii)            =|5x1−12y1+2625+144|=|5x1−12y1+2613|  of  (x1,y1)  from  eqn.(ii)            =|7x1+24y1−5049+576|=|7x1+24y1−5025|If  the    (x1,y1)  is    from  the  given  lines,  then|4x1+3y1+105|=|5x1−12y1+2613|=|7x1+24y1−5025|We  see  that  putting  x1=0  and  y1=0,  the  above  relation  is  satisfied  i.e.,        105=2613=5025=2Hence,  the  correct  option  is  (c).

New answer posted

a year ago

Class 11th Exemplar Solution

The point (4, 1) undergoes the following two successive transformations:

i. Reflection about the line y = x

ii. Translation through a distance 2 units along the positive x-axis

Then the final coordinates of the point are

(a) (4, 3)

(b) (3, 4)

(c)  (1, 4)

(d)   7 2 , 7 2  

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

Let  the  reflection  of  A(4,1)  in  y=x  be  B(a,b)  mid  of  AB=(4+a2,1+b2)which  lies  on  y=x⇒          4+a2=1+b2⇒           4+a=1+b⇒           a−b=−3                                           …(i)The  slope  of  the  line  y=x  is  1  and  slope  of  AB=b−1a−4∴      1(b−1a−4)=−1⇒              b−1=−a+4⇒             a+b=5                                           …(ii)Solving  eqn.(i)  and  eqn.(ii)  we  get              a=1  and  b=4∴  The    after  translation  is  (1+2,4)  or  (3,4)Hence,  the  correct  option  is  (b).

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