1. P(1,2), Q(25,2), R(10,20).
X = (1+25+10) / 3 = 12.
Y = (2+2+20) / 3 = 8.
C(12,8). = Coordinates of Centroid.
2. SIDE PQ:
Xo = (1+25) / 2 = 13.
Yo = (2+2) / 2 = 2.
M(13,2).
m = (2-2) / (25-1) = 0/24 = 0.
m2 = -1/0 = Undefined = slope of perpendicular bisector(P.B.).
Eq1: X = 13 = Eq of P.B. of PQ = a ver line.
SIDE PR:
Xo = (1+10) / 2 = 5.5.
Yo = (2+20) / 2 = 11.
M(5.5,11).
m = (20-2) / (10-1) = 2.
m2 = -1/2 = slope of P.B. of PR.
Y = (-1/2)*5.5 + b = 11,
-2.75 + b = 11,
b = 11 + 2.75 = 13.75.
Eq2: Y = -X/2 + 13.75 = P.B. OF PR.
Substitute 13 for X:
Y = -13/2 + 13.75 = 7.25.
C(13,7.25).=Coordinates of circumcenter
Triangle PQR has vertices P(1,2), Q(25,2) and R(10,20).
Find the coordinates of the centroid.
Find the coordinates of the circumcenter.
Find the coordinates of the orthocenter.
Find the equation of the line.
1 answer
1 answer