Question

The equation to the locus of points equidistant from the points (2,3),(-2,5) is..........

Answer 1

Step-by-step explanation:

If the equation to the locus of points equidistant from the points (−2,3),(6,−5) is ax+by+c=0 where a>0 then, the ascending order of a,b,c is.

Answer 2

Answer:

2x + y - 4 = 0

Step-by-step explanation:

The distance between the points P (x1, y1)and Q (x2, y2) is

PQ = √(x2 − x1) 2 + (y2 − y1)2 , which is called the distance formula.

Let P (x, y) be the locus of points equidistant from the points (2, 3) & (-2, 5)

Let A ( 2,3) & B( -2, 5)

Now,

PA = PB

or PA² = PB²

Distance formula is

d = √(x2 - x1)² -  (y2 - y1)²

Applying the formula

(2 - x)² - (3 - y)² = (-2 - x)² - (5 - y)²

Applying the identity [(a -b)² = a² - 2ab + b²] to all the brackets

(4 - 4x + x²) - (9 - 6y + y²) = (4 + 4x + x²) - (25 - 10y + y²)

4 - 4x + x² - 4 - 4x -x²  -9 + 6y - y² +25 - 10y + y² = 0

-8x - 4y + 16 = 0

Or, 8x + 4y - 16 = 0

Dividing the equation by 4

Ans: 2x + y - 4 = 0