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The image of the point (3, 5) in the line x – y + 1 = 0, lies on:

Option 1 -

( x − 4 ) 2 + ( y + 2 ) 2 = 1 6

Option 2 -

( x − 2 ) 2 + ( y − 4 ) 2 = 4

Option 3 -

( x − 2 ) 2 + ( y − 2 ) 2 = 1 2

Option 4 -

( x − 4 ) 2 + ( y − 4 ) 2 = 8

0 3 Views | Posted a month ago
Asked by Shiksha User

  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    a month ago
    Correct Option - 4


    Detailed Solution:

    Equation of given line x – y + 1 = 0 .(i),

    equation of per perpendicular line PP’ is

    -x – y + λ = 0  

    As line passing through (3, 5)

    ∴ − 3 − 5 + λ = 0   

    λ = 8  

    Equation of line PP’ is –x – y + 8 = 0 .(ii)

    Solving (i) and (ii)

    − 2 y = − 9 y = 9 2 } Q = ( 7 2 , 9 2 )

    y = 9 2

    ∴ x − 9 2 + 1 = 0

    P ' = ( 4 , 4 )           

    5 + y 1 2 = 9 2           

    y1 = 4

     

Similar Questions for you

A line passes through (9, 0), making angle 30° with positive direction of x-axis. It is rotated by angle of 15° with respect to (9, 0). Then one of the equation of new line is
A
alok kumar singh

Eqn : y – 0 = tan45° (x – 9) Þ y = (x – 9)

Option (B) is correct

If two circle x2 + y2 = 4 and x2 + y2 – 4lx + 9 = 0 intersect at two distinct points, then find the range of l.
A
alok kumar singh

|r1 – r2| < c1c2 < r1 + r2

-> | 2 − 4 λ 2 − 9 | < | 2 λ | < 2 + 4 λ 2 − 9

| 2 λ | − 2 < 4 λ 2 − 9    

4 λ 2 + 4 − 8 | λ | < 4 λ 2 − 9

  λ > 1 3 8 ,   λ < − 1 3 8           

4 λ 2 − 9 > 0

λ > 3 2 ,   λ < − 3 2

λ ∈ ( − ∞ ,   − 1 3 8 ) ∪ ( 1 3 8 ,   ∞ )           

Now,

| 2 − 4 λ 2 − 9 | < | 2 λ |            

4 + 4 λ 2 − 9 − 4 4 λ 2 − 9 < 4 λ 2  

4 4 λ 2 − 9 > − 5 ⇒ λ ∈ R  

λ ∈ ( − ∞ ,   − 1 3 8 ) ∪ ( 1 3 8 ,   ∞ )  

The line passes through the centre of circle x2 + y2 – 16x – 4y = 0, it interacts with the positive coordinate axis at A & B. Then find the minimum value of OA + OB, where O is origin.
A
alok kumar singh

(y – 2) = m (x – 8)

⇒   x-intercept

⇒     ( − 2 m + 8 )

⇒   y-intercept

⇒   (–8m + 2)

⇒   OA + OB = − 2 m 2  + 8 – 8m + 2

f ' ( m ) = 2 m 2 − 8 = 0  

-> m 2 = 1 4

-> m = − 1 2

-> f ( − 1 2 ) = 1 8

->Minimum = 18

The image of the point (3, 5) in the line x – y + 1 = 0, lies on:
V
Vishal Baghel

Kindly consider the following figure

A man is walking on a straight line. The arithmetic mean of the reciprocals of the intercepts of the line on the coordinate axes is 1 4 .  Three stones A, B and C are placed at the point (1, 1), (2, 2) and (4, 4) respectively. Then which of these stones is / are on the path of the man?
V
Vishal Baghel

According to question,

1 2 ( 1 a + 1 b ) = 1 4

1 a + 1 b = 1 2 . . . . . . . . ( i )

Equation of required line is x a + y b = 1

Obviously B (2, 2) satisfying condition (i)

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