Mathematics - Online Test

Consider any a ,b , c ∈$\in$A .

1. Since both a and a must be either even or odd, so (a , a) ∈R ⇒R is reflexive.
2. Let (a ,b) ∈R ⇒ both a and b must be either even or odd, ⇒ both b and a must be either even or odd, ⇒ ( b ,a) ∈R .Thus , (a ,b) ∈R ⇒ ( b ,a) ∈R ⇒R is symmetric.
3. Let (a ,b) ∈R and (b ,c) ∈R ⇒ both a and b must be either even or odd, also ,both b and c must be either even or odd, ⇒ all elements a, b and c must be either even or odd, ⇒ ( a ,c) ∈R . Thus , (a ,b) ∈R ⇒ ( b ,c) ∈R ⇒ (a ,c) ∈R ⇒R is transitive.

The equation of the pair of straight lines paasing through the origin is fiven by is given by ax2 + 2hxy + by2 =0

The condition for perpendicularity between the pair of lines is a+b = 0

Hence the locus of the pair of these perpendicular straight lines should be xy = 0

p∨∼p≡T

T∨p≡T

Because line joining the points (a, 0) and (0, b)  will be diameter and it's mid-point will be centre of the circle.

Or let the equation of circle be (x−h)2+(y−k)2=r2

as circle passes through  (0,0) , (a,0) and (0,b), they will satisfy the above equation

putting the points in above equation we get

h2+k2=r2

(a−h)2+k2=r2  ---(ii)

h2+(b−k)2=r2 ---(iii)

Solving (i) and (ii),we get  h=a2

whereas solving (i) and (iii) we get k=b2

hence its center is (a2,b2)