State True or False.Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be”. When this situation is represented algebraically and graphically, we will observe that the lines will be parallel to each other.

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Accepted Answer

Given that seven years ago, Aftab was seven times as old as his daughter and also three years from now, he will be three times as old as his daughter will be. Let the present age of Aftab be x and the present age of his daughter be y.Seven years ago,Age of Aftab = x – 7,Age of his daughter = y – 7.Aftab was seven times as old as his daughter, seven years ago.So algebraically, it will become

x−7=7
      
       y−7

⇒x−7=7y−49

⇒x−7y=−42

⇒x−7y+42=0   ------(1)

Also, given that after three years from now,Then the age of Aftab

=
    
     x+3
   
  ,and the age of his daughter

=
    
     y+3
   
  .Now it is also given that the age of Aftab after three years will become three times that of his daughter. So,

x+3=3
      
       y+3

⇒x+3=3y+9

⇒x−3y=6

⇒x−3y−6=0   ------(2)

We know that the pair of linear equations in two variables x and y is represented by the standard form,

a
      1
     
     x+
      b
      1
     
     y+
      c
      1
     
     =0

a
      2
     
     x+
      b
      2
     
     y+
      c
      2
     
     =0

Here

a
    1
   
   ,
    a
    2
   
   ,
    b
    1
   
   ,
    b
    2
   
   ,
    c
    1
   
   ,
    c
    2

are the coefficients of the equations.On comparing the ratios

a
      1

a
      2

,

b
      1

b
      2

,

c
      1

c
      2

, we can determine whether the lines representing the given pairs of linear equations intersect at a point, are parallel or coincide.For the given pair of linear equations to be inconsistent or the lines representing them to be parallel, the comparison of

a
      1

a
      2

,

b
      1

b
      2

and

c
      1

c
      2

must satisfy the condition

a
      1

a
      2

=

b
      1

b
      2

≠

c
      1

c
      2

.
  Now, let us check the condition for parallel lines in the following equations.

x−7y+42=0
  ,

x−3y−6=0
  .Here,

a
      1
     
     =1,
      b
      1
     
     =−7,
      c
      1
     
     =42,

a
      2
     
     =1,
      b
      2
     
     =−3,
      c
      2
     
     =−6.

So,

a
        1

a
        2

=
      1
      1

b
        1

b
        2

=
      
       −7
      
       −3

⇒

a
        1

a
        2

≠

b
        1

b
        2

Since

a
      1

a
      2

≠

b
      1

b
      2

, the lines are not parallel.Now let us plot a graph for the two equations.For the equation x – 3y – 6 = 0, find the x and y values.x06y

−2
  0 For the equation: x – 7y + 42 = 0, find the x and y values.x0

−42
  y

6
  0On plotting the lines using the above tables, we get,As we can observe clearly from graph that the lines intersect at (42, 12).Hence the statement is false.Therefore, when the given situation is represented algebraically and graphically, we will observe that the lines will be intersecting.So, option 2 is correct.

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x	0	6
y	$− 2$	0

x	0	$− 42$
y	$6$	0