Find the coordinates of the point equidistant from the points A(1,2),B(3,-4)and C(5,-6).
Answers
Step-by-step explanation:
The three given points are A(1, 2), B(3, -4), and C(5, -6).
Let P (x, y) be the point that is equidistant from the given three points.
So, PA = PB = PC ---- (1)
According to distance formula, we have distance between any two points (x, y) and (
x
1
,
y
1
) = √(x -
x
1
)2 + (y -
y
1
)2.
Distance PA = √(x - 1)2 + (y - 2)2
Distance PB = √(x - 3)2 + (y + 4)2
Distance PC = √(x - 5)2 + (y + 6)2
Substituting these in equation (1),
√(x - 1)2 + (y - 2)2 = √(x - 3)2 + (y + 4)2 = √(x - 5)2 + (y + 6)2
⇒ x2 + 1 - 2x + y2 + 4 – 4y = x2 + 9 - 6x + y2 + 16 + 8y = x2 + 25 – 10x + y2 + 36 + 12y
⇒ 1 – 2x + 4 – 4y = 9 - 6x + 16 + 8y = 25 – 10x + 36 + 12y
Considering, 1 – 2x + 4 – 4y = 9 - 6x + 16 + 8y, we get,
⇒ 2x - 6y = 10 … (1)
Now, consider, 1 – 2x + 4 – 4y = 25 – 10x + 36 + 12y
⇒ x – 2y = 7 … (2)
Solving (1) and (2), we get x = 11, y = 2.
Thus, the required point is (11, 2).
We gat the equation 1and2
2x-6y=10
x-2y=7
(11,2)is the ans of the equation