Question

The point a (2, 1) is translated parallel to the line x - y = 3 by a distance 4 units. If the new- position a' is in third quadrant, then the co-ordinates of a' are

Answer 1

Given:

A point a(2, 1) is translated parallel to the line x - y = 3 by a distance 4 units.

New- position a' is in third quadrant.

To Find:

Co-ordinates of a' .

Solution:

Given that a(2, 1) is translated parallel to the line x - y = 3 .

Therefore a line parallel to x - y =3 is x - y = k , where k is some constant.

• Here x  - y = k passes through ( 2, 1 ) .

Therefore ,

• k = 2 -1 = 1 .

Hence ( 2, 1) is translated through x - y = 1 .

Let new position of a ,  

• a' = (  p, q ).
• ( p, q ) will lie on x - y = 1

Hence,

• p - q = 1 = = > p = q + 1

Also its in 3rd quadrant , hence both will be negative.

• Given ( p, q) is at a distance of 4 units from ( 2, 1)

By distance formula ,

• $\sqrt{(p - 2)^{2} + (q -1)^{2} }$ = 4
• $\sqrt{( q + 1 - 2 )^{2} + ( q - 1 )^{2} }$ = 4
• ( q - 1 )² + (q - 1 )² = 4²
• 2 ( q - 1 )² = 4²
• ( q -1 )² = 8
• q - 1 = ±2√2
• q = ±2√2 + 1

Hence,

• q = -2√2 + 1 =  -1.83, as q is in 3rd quadrant.
• p = -2√2 + 2 = -0.83

Therefore ,

the co-ordinates of a' are ( -0.83, - 1.83 )