1. The equation of the locus of a point which is always equidistant from the points (a + b, a - b) and (a - b, a + b) is a) x - a=0 b) x - y = 0 c) y + x = 0 d) x + y = a + b

c). y + x = 0

Step-by-step explanation:

The locus is a line perpendicular to the line connecting the 2 points and passing through the centre of the line.

The mid-point of the line connecting the given points is:

(a + b + a - b)/2 , (a - b + a + b) / 2

= (a, a)

It's slope = (a + b)- (a - b) / (a - b)- (a + b)

= 2b/ -2b

= -1

So the slope of the line we want is -1/(-1) = 1.

Using the point-slope form of a line:

y - y1 = m(x - x1)

y - a = 1(x - a)

y - a = x - a

x + y = 0

f(x) = x^3 -2x^2 -3x

x( x^2 -2x-3)

x( x^2 +x -3x -3)

x( x(x+1) -3(x+1))

x( x-3)( x+1)

so zeroes are 0, 3,-1

its (2) option