Let A(4,2) ; B(6,7) ;C(1,4) be the vertices of triangle ABC,to find [i]The median from A meets BC at D,To - Maths - Coordinate Geometry

Question

Let A(4,2) ; B(6,7) ;C(1,4) be the vertices of triangle ABC,to
find
[i]The median from A meets BC at D,To Find co-ordinates of D
[ii]Find the co-ordinates of the point P on AD such that AP:PD =
2:1
[iii]Find the co-ordinates of point Q and R on medians BE and
CFrespectively such that BQ:QE = 2:1
[iv] what do you observe ?
pls help

Utsav · Accepted Answer

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As AD is the median of the triangle so D is the midpoint of BC
Co-ordinates of D using midpoint formula is
D=(6+12),(7+42)=72,112Now it is given AP:PD = 2:1 A(4,2) P(x,y)
D72,112Hence coordinates of P using section formula is
P=2×72+1×42+1,2×112+1×22+1∴Coordinates
of P is (113,133)Now BE is also a median so E is the midpoint of AC
Co-ordinates of E using midpoint formula is E=(4+12),(2+42)=52,3Now
it is given BQ:QE = 2:1 B(6,7) Q(x,y) E52,3Hence coordinates of Q
using section formula is
Q=2×52+1×62+1,2×3+1×72+1∴Coordinates
of Q is (113,133)Now CF is also a median so F is the midpoint of AB
Co-ordinates of F using midpoint formula is F=(4+62),(2+72)=5,92Now
it is given CR:RF = 2:1 C(1,4) R(x,y) F5,92Hence coordinates of R
using section formula is
R=2×5+1×12+1,2×92+1×42+1∴Coordinates
of R is (113,133)

It is observed that points P,Q and R are having the same
co-ordinates Hence that point is the centroid where medians of a
triangle meet Hence we can say that centroid divides the median in
the ratio 2:1