First slide

Cartesian plane

Question

If an triangle $ABC, A \equiv (1, 10),$ circumcentre $\equiv (- 1 / 3, 2 / 3),$ and orthocentre $\equiv (- 11 / 3, 4 / 3),$ then the coordinates of the midpoint of the side opposite to A are

Moderate

A

(1) $(1, - 11 / 3)$
B

(2) $(1, 5)$
C

(3) $(1, - 3)$
D

(4) $(1, 6)$

Solution

Circumcentre $O \equiv (- 1 / 3, 2 / 3)$ and orthocenter $H \equiv (11 / 3, 4 / 3)$
Therefore, the coordinates of G are (1,8/9). Now, the point A is (1,10) as G is (1/ 8/9).
Also, AD : DG =3:1
$∴ D \equiv (\frac{3 - 1}{3 - 1}, \frac{8 / 3 - 10}{3 - 1}) \equiv (1, 11 / 3)$
$∴ D = (\frac{3 - 1}{2}) = 1, D_{y} = \frac{(8 / 3) - 10}{2} = - \frac{11}{3}$
Hence, the coordinates of t'he midpoint of BC are (1,-11 /3).

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