Question

In a triangle ABC , co-ordinates of A are $(1,2)$ and the equations of the medians through B and C are $x+y=5$ and $x=4$ respectively. Then the coordinates of B and C will be

• A. $(-2,7),(4,3)$
• B. $(7,-3),(4,3)$
• C. $(2,7),(-4,3)$
• D. $(2,-7),(4,-3)$

Answer 1

The correct option is B $(7,-3),(4,3)$

$\Delta ABC$ has vertex $A(1,2)$

Given that midpoint of $AC$ is $D$ and mid point of $AB$ is $E$

Also,

equation of medium $BD$ is $x+y=5$

equation of medium $CE$ is $x=4$

let coordinate of $B(p,q)$ and $C(r,s)$

$\therefore$ coordinate of $E\equiv (\frac{p+1}{2},\frac{q+1}{2})$

But $E$ lies on the line segment $CE$ $(x=4)$

$\therefore$ $\frac{P+1}{2}=4$

$\Rightarrow P=7$

And point $B(p,q)$ lies on the line $BD$ $[x+y=4]$

$\therefore$ $p+q=4$

$\Rightarrow q=-3$

$\therefore$ $B\equiv (7,-3)$

Point $C(r,s)$ lies on $CE$ $[x=4]$

$\therefore$ $r=4$

Coordinates of $D$ are $\equiv [\frac{r+1}{2},\frac{s+2}{2}]$

Similarly point $D$ lies on $BD$ $[x+y=5]$

$\therefore$ $\frac{r+1}{2}+\frac{s+2}{2}=5$

$r+s=7$

$\Rightarrow s=3$

$\therefore$ $C\equiv (4,3)$

$\therefore$ coordinates of $B$ & $C$ are $(7,-3)$ & $(4,3)$