The two points on the line `x +y =4 `that at a unit perpendicular distance from the line lie `4x +3y =10` are `(a_1, b_1)` and `(a_2, b_2)`, then `a_1+ b_1+a_2+ b_2 `

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 ?

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

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 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)

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 find the value of c .

If the points (a_1, b_1),\ \ (a_2, b_2) and (a_1+a_2,\ b_1+b_2) are collinear, show that a_1b_2=a_2b_1 .

If the points (a_1, b_1),\ \ (a_2, b_2) and (a_1+a_2,\ b_1+b_2) are collinear, show that a_1b_2=a_2b_1 .