`y^2=21-x`
`x=5`
The solutions to the system of equation above are `(a_1,b_1) and (a_2,b_2)`. What are the values of `b_1` and `b_2`?

{(y=(1)/(2)x-2), (y=-x^(2)+1):} If (a, b) is a solution to the system of equations above, which of the following could be the value of b?

If a_1/a_2!=b_1/b_2 then the system of equation a_1x+b_1y+c_1=0 , a_2x+b_2y+c_2=0 has

In direct proportion a_1/b_1 = a_2/b_2

If the equation of the locus of a point equidistant from the points (a_1,b_1) and (a_2,b_2) is (a_1-a_2)x+(b_1-b_2)y+c=0 , then the value of C is

If the equation of the locus of a point equidistant from the points (a_1,b_1) and (a_2,b_2) is (a_1-a_2)x+(b_2+b_2)y+c=0 , then the value of C is

If the equation of the locus of a point equidistant from the points (a_1,b_1) and (a_2,b_2) is (a_1-a_2)x+(b_2+b_2)y+c=0 , then the value of C is

If the equation of the locus of a point equidistant from the points (a_1, b_1) and (a_2, b_2) is (a_1-a_2)x+(b_1-b_2)y+c=0 , then the value of c is

If the equation of the locus of a point equidistant from the points (a_1, b_1) and (a_2, b_2) is (a_1-a_2)x+(b_1-b_2)y+c=0 , then the value of c is a a2-a2 2+b1 2-b2 2 sqrt(a1 2+b1 2-a2 2-b2 2) 1/2(a1 2+a2 2+b1 2+b2 2) 1/2(a2 2+b2 2-a1 2-b1 2)

If the equation of the locus of a point equidistant from the points (a_1, b_1) and (a_2, b_2) is (a_1-a_2)x+(b_1-b_2)y+c=0 , then the value of c is a a2-a2 2+b1 2-b2 2 sqrt(a1 2+b1 2-a2 2-b2 2) 1/2(a1 2+a2 2+b1 2+b2 2) 1/2(a2 2+b2 2-a1 2-b1 2)