question is: find the coordinates of midpoint P of line BC. Calculate lengths of AP, BP, CP. P is ? (midpoint of BC which bisects <A(0,0), B(0, 2k), C(2h,0). Thank you.
the midpoint is the average of the endpoints.
So, for BC, P=(h,k)
Now just use your distance formula to get the required lengths.
To find the coordinates of the midpoint P of line BC, we need to calculate the average of the x-coordinates and the average of the y-coordinates of points B and C.
Given points A(0,0), B(0, 2k), C(2h,0), we can calculate the coordinates of P as follows:
1. Calculate the midpoint coordinates:
- The x-coordinate of P is the average of the x-coordinates of B and C:
P_x = (B_x + C_x) / 2 = (0 + 2h) / 2 = h
- The y-coordinate of P is the average of the y-coordinates of B and C:
P_y = (B_y + C_y) / 2 = (2k + 0) / 2 = k
Therefore, P has coordinates (h, k).
2. Calculate the lengths of AP, BP, and CP:
- The distance between two points can be calculated using the distance formula:
Distance = sqrt[(x2 - x1)^2 + (y2 - y1)^2]
- AP: (0, 0) to (h, k)
AP_length = sqrt[(h - 0)^2 + (k - 0)^2] = sqrt[h^2 + k^2]
- BP: (0, 2k) to (h, k)
BP_length = sqrt[(h - 0)^2 + (k - 2k)^2] = sqrt[h^2 + (-k)^2] = sqrt[h^2 + k^2]
- CP: (2h, 0) to (h, k)
CP_length = sqrt[(h - 2h)^2 + (k - 0)^2] = sqrt[(-h)^2 + k^2] = sqrt[h^2 + k^2]
Therefore, the lengths of AP, BP, and CP are all sqrt[h^2 + k^2].
In summary, the coordinates of the midpoint P of line BC are (h, k), where h is the x-coordinate average of B and C, and k is the y-coordinate average of B and C. The lengths of AP, BP, and CP are all sqrt[h^2 + k^2].