Vedanta Excel In Opt. Mathematics - Book 9 Find the locus of a point which moves so that

(a) Its distance from (1, 2) and (2, -3) are equal.​

(1 , 2) (2, -3) (x1,y1) (x2, y2) Distance formual √(x2-x1)+(y2-y1) {substitute numbers in formula} = √(2-1)^+(-3-2)^ =√(1)^+(-5)^ [here square root and square get cancelled] = (1)+(-5) = 1-5 = -4

A point that is equidistant from two points will travel along the perpendicular bisector of the line that joins them.
The gradient of the line between (1,2) and (2,-3) is (-3 - 2)/(2 - 1) = -5
The perpendicular gradient = -1/-5 or 0.2
The midpoint between (1,2) and (2,-3) is ((1+2)/2,(2+-3)/2) = (1.5, -0.5)
Substituting into y=mx + c
-0.5 = 0.2 x 1.5 + c
-0.5 = 0.3 + c
c = -0.5 - 0.3 = -0.8
So the locus is the line y = 0.2x - 0.8

Sometimes life is like a tunnel. you can’t always see the light at the end of the tunnel, but if you keep moving, you will come to a better place