(i) The median from A meets BC in D. Find the coordinates of the point D.
Here given vertices are A (4, 2), B (6, 5) and C (1, 4).
By midpoint formula.
x = \(\frac{x_1+x_2}2\), y = \(\frac{y_1+y_2}2\)
For midpoint D of side BC,
x = \(\frac{6+1}2\), y = \(\frac{5+4}2\)
x = \(\frac{7}2\), y = \(\frac{9}2\)
Hence, the coordinates of D are (\(\frac{7}2\),\(\frac{9}2\))
(ii) Find the coordinates of point P on AD such that AP : PD = 2 :1.
By section formula,
x = \(\frac{mx_2+nx_1}{m+n}\), y = \(\frac{my_2+ny_1}{m+n}\)
For point P on AD, where m = 2 and n = 1
(iii) Find the coordinates of the points Q and R on medians BE and CF respectively such that BQ : QE = 2 : 1 and CR : RF = 2 : 1.
By midpoint formula.
x = \(\frac{x_1+x_2}2\), y = \(\frac{y_1+y_2}2\)
For midpoint E of side AC,
x = \(\frac{1+4}2\), y = \(\frac{4+2}2\)
x = \(\frac{5}2\), y = \(\frac{6}2\)
Hence, the coordinates of E are ( \(\frac{5}2\), 3)
For midpoint F of side AB,
x = \(\frac{6+4}2\), y = \(\frac{5+2}2\)
x = \(\frac{10}2\), y = \(\frac{7}2\)
Hence, the coordinates of F are (5, \(\frac{7}2\))
By section formula,
x = \(\frac{mx_2+nx_1}{m+n}\), y = \(\frac{my_2+ny_1}{m+n}\)
For point Q on BE, where m = 2 and n = 1
For point R on CF, where m = 2 and n = 1
(iv) What do you observe?
We observe that the point P,Q and R coincides with the centroid.
This also shows that centroid divides the median in the ratio 2:1
