Find the locus of the point equidistant from (a+b b-a) and (a-b a+b)
Answers
Answer:
bx - ay = 0
Step-by-step explanation:
let A = (a+b , b-a)
B = ( a-b , a+b)
let P(h,k) be the locus of the point equidistant from A and B.
Distance formula :- distance between two points = √(x₂ - x₁)² - ( y₂- y₁)²
PA = PB
PA² = PB²
((a+b) -h)² + ((b-a)-k)² = ((a-b)-h)² + ((a+b)-k)²
(a+b)²+ h² -2h(a+b) +(b-a)² + k²- 2k(b-a) =(a-b)²+h²-2h(a-b)+(a+b)²+ k - 2k(a+b)
2h (a+b) + 2k (b-a) = 2h(a-b) + 2k(a+b)
ah + hb + bk -ak = ah -hb + ak + bk
2bh = 2ak
bh = ak
substituting h and k with x and y
bx = ay
bx - ay =0
the locus of the point equidistant from (a+b , b-a) and ( a-b, a+b) is bx-ay=0