Radius of circumcircle of a triangle ABC is units. If point P is equidistant from A (1, 3), B(-3, 5) and C(5,-1) then AP
Question
Radius of circumcircle of a triangle ABC is units. If point P is equidistant from A (1, 3), B(-3, 5) and C(5,-1) then AP
Solution
To find the length of AP, we can use the distance formula. The distance formula states that the distance between two points (x1, y1) and (x2, y2) is given by the formula:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Let's calculate the distances between point P and points A, B, and C.
Distance between P and A: d1 = sqrt((1 - x)^2 + (3 - y)^2)
Distance between P and B: d2 = sqrt((-3 - x)^2 + (5 - y)^2)
Distance between P and C: d3 = sqrt((5 - x)^2 + (-1 - y)^2)
Since point P is equidistant from points A, B, and C, we can set up the following equations:
d1 = d2 d1 = d3
Substituting the values of d1, d2, and d3, we get:
sqrt((1 - x)^2 + (3 - y)^2) = sqrt((-3 - x)^2 + (5 - y)^2) sqrt((1 - x)^2 + (3 - y)^2) = sqrt((5 - x)^2 + (-1 - y)^2)
Squaring both sides of the equations, we get:
(1 - x)^2 + (3 - y)^2 = (-3 - x)^2 + (5 - y)^2 (1 - x)^2 + (3 - y)^2 = (5 - x)^2 + (-1 - y)^2
Expanding and simplifying the equations, we get:
1 - 2x + x^2 + 9 - 6y + y^2 = 9 + 6x + x^2 + 25 - 10y + y^2 1 - 2x + x^2 + 9 - 6y + y^2 = 25 - 10x + x^2 + 1 + 2y + y^2
Simplifying further, we get:
-2x - 6y + 10x - 2y = 25 - 1 - 9 8x - 8y = 15
Dividing both sides by 8, we get:
x - y = 15/8
So, the equation of the line passing through points A(1, 3) and P(x, y) is x - y = 15/8.
Now, let's find the coordinates of the circumcenter of triangle ABC. The circumcenter is the point of intersection of the perpendicular bisectors of the sides of the triangle.
To find the equation of the perpendicular bisector of side AB, we need to find the midpoint of AB and the slope of AB.
Midpoint of AB: x_mid = (1 + (-3))/2 = -1/2 y_mid = (3 + 5)/2 = 4/2 = 2
Slope of AB: m_AB = (5 - 3)/(-3 - 1) = 2/-4 = -1/2
The equation of the perpendicular bisector of AB can be found using the point-slope form:
y - y_mid = -1/m_AB * (x - x_mid)
Substituting the values, we get:
y - 2 = -1/(-1/2) * (x - (-1/2)) y - 2 = 2 * (x + 1/2) y - 2 = 2x + 1 y = 2x + 3
Similarly, we can find the equations of the perpendicular bisectors of sides BC and AC:
Perpendicular bisector of BC: y = -2x + 1
Perpendicular bisector of AC: y = 1/2x - 1/2
The circumcenter is the point of intersection of these three lines. To find the coordinates of the circumcenter, we can solve the system of equations formed by these three lines:
x - y = 15/8 y = 2x + 3 y = -2x + 1
Substituting the second and third equations into the first equation, we get:
2x + 3 - (-2x + 1) = 15/8 2x + 3 + 2x - 1 = 15/8 4x + 2 = 15/8 4x = 15/8 - 2 4x = 15/8 - 16/8 4x = -1/8 x = -1/32
Substituting the value of x into the second equation, we get:
y = 2(-1/32) + 3 y = -1/16 + 48/16 y = 47/16
So, the coordinates of the circumcenter are (-1/32, 47/16).
Now, we can find the distance between point A(1, 3) and the circumcenter (-1/32, 47/16) using the distance formula:
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