if A=(3,5) and B=(4,1).find the equation of locus of point p which are equidistant from points A and B
such a point must lie on the right-bisector of AB
midpoint of AB =(7/2 , 3)
slope of AB = (1-5)/(4-3) = -4
so slope of perpendicular = 1/4
equation:
y-3 = (1/4)(x-7/2)
y - 3 = (1/4)x - 7/8
y = (1/4)x +3 - 7/8
y = (1/4)x + 17/8
To find the equation of the locus of points equidistant from points A and B, we need to find the midpoint of the line segment AB, and then find the distance from this midpoint to any point P.
Let's first find the midpoint of AB.
The midpoint formula states that the coordinates of the midpoint (M) of a line segment with endpoints (x1, y1) and (x2, y2) are given by:
M = ((x1 + x2) / 2, (y1 + y2) / 2)
Using this formula, we can find the midpoint of AB as follows:
M = ((3 + 4) / 2, (5 + 1) / 2)
M = (7 / 2, 6 / 2)
M = (7/2, 3)
Now, let's find the distance between the midpoint M and any point P(x, y) using the distance formula.
The distance formula is:
d = √((x2 - x1)^2 + (y2 - y1)^2)
Applying this formula, the distance from M to any point P(x, y) is:
d = √((x - 7/2)^2 + (y - 3)^2)
Since we want the locus of points equidistant from A and B, the distance from M to any point P should be equal to the distance from A to B.
Let's calculate the distance between A and B:
dAB = √((4 - 3)^2 + (1 - 5)^2)
dAB = √(1 + 16)
dAB = √17
The equation of the locus will be:
√((x - 7/2)^2 + (y - 3)^2) = √17
Squaring both sides of the equation, we get:
(x - 7/2)^2 + (y - 3)^2 = 17
Therefore, the equation of the locus of points equidistant from A(3, 5) and B(4, 1) is (x - 7/2)^2 + (y - 3)^2 = 17.