Question

A(-3;P) is equidistant from the points C(7;-1) and D(4;-4).
Find the value of p.

Answer 1

To find the value of p, we first need to determine the distance between the points A and C, and the distance between the points A and D. Since A is equidistant from C and D, the distances will be equal.

The distance between two points (x1, y1) and (x2, y2) can be found using the distance formula:

Distance = √((x2 - x1)^2 + (y2 - y1)^2)

For the distance between A(-3, p) and C(7, -1):

Distance_AC = √((7 - (-3))^2 + (-1 - p)^2)
Distance_AC = √(10^2 + (-1 - p)^2)
Distance_AC = √(100 + 1 - 2p + p^2)
Distance_AC = √(101 - 2p + p^2)

For the distance between A(-3, p) and D(4, -4):

Distance_AD = √((4 - (-3))^2 + (-4 - p)^2)
Distance_AD = √(7^2 + (-4 - p)^2)
Distance_AD = √(49 + 16 + 8p + p^2)
Distance_AD = √(65 + 8p + p^2)

Since A is equidistant from C and D, we have:

√(101 - 2p + p^2) = √(65 + 8p + p^2)

Squaring both sides to eliminate the square root, we get:

101 - 2p + p^2 = 65 + 8p + p^2
101 - 2p = 65 + 8p
36 = 10p
p = 3.6

Therefore, the value of p is 3.6.