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 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 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 .

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 ?

The ratio in which the line segment joining points A(a_1,\ b_1) and B(a_2,\ b_2) is divided by y-axis is (a) -a_1: a_2 (b) a_1: a_2 (c) b_1: b_2 (d) -b_1: b_2

If the origin lies in the acute angle between the lines a_1 x + b_1 y + c_1 = 0 and a_2 x + b_2 y + c_2 = 0 , then show that (a_1 a_2 + b_1 b_2) c_1 c_2 lt0 .

If the tangent and the normal to x^2-y^2=4 at a point cut off intercepts a_1,a_2 on the x-axis respectively & b_1,b_2 on the y-axis respectively. Then the value of a_1a_2+b_1b_2 is equal to:

If A_1,B_1,C_1, , are respectively, the cofactors of the elements a_1, b_1c_1, , of the determinant "Delta"=|a_1b_1c_1a_1b_2c_2a_3b_3c_3|,"Delta"!=0 , then the value of |B_2C_2B_3C_3| is equal to a1^2Δ b. a_1Δ c. a_1^2Δ d. a1 2^2Δ

If the tangent and normal to xy=c^2 at a given point on it cut off intercepts a_1, a_2 on one axis and b_1, b_2 on the other axis, then a_1 a_2 + b_1 b_2 = (A) -1 (B) 1 (C) 0 (D) a_1 a_2 b_1 b_2

If curves a_1 x^2 + 2h_1 xy + b_1 y^2 - 2g_1 x - 2f_y y + c = 0 and a_2 x^2 - 2h_2xy + (a_2 + a_1 - b_1) y^2 - 2g_2 x - 2f_2 y + c = 0 intersect in four concyclic points A, B, C and D . and H be the point ((g_1+g_2)/(a_1 + a_2), (f_1 + f_2)/(a_1 + a_2)) , then (5HA^2 + 7HB^2 + 8HC^2)/(HD^2) = ..