In a triangle ABC, coordinates of A are (1,2) and the equations of the medians through B a…

Question

In a triangle $ABC$, coordinates of $A$ are $(1,2)$ and the equations of the medians through $B$ and $C$ are $x+y=5$ and $x=4$ respectively. Then area of $	riangle ABC$ (in sq. units) is:

Accepted Answer

Since $C$ lies on the median through $C$... wait --- the median through $B$ passes through the midpoint of $AC$, and the median through $C$ (line $x=4$) means $C$ has $x=4$, and the midpoint $D$ of $AC$ lies on the median through $B$ ($x+y=5$).

$C=(4,y_C)$; midpoint of $A(1,2),C(4,y_C)$ is $D=\left(\dfrac52,\dfrac{2+y_C}2\right)$, lying on $x+y=5$:
$$\frac52+\frac{2+y_C}2=5 \implies 2+y_C=5 \implies y_C=3$$

So $C=(4,3)$.

Centroid $G$ = intersection of medians $x=4$ and $x+y=5$: $G=(4,1)$.

$$\text{Area}(\triangle ABC) = 3\times\text{Area}(\triangle AGC) = 3\times\frac12|1(1-3)+4(3-2)+4(2-1)| = 3\times\frac12|{-2+4+4}| = 3\times3=9$$

Solutions · 1