Given the point $A(0,-1)$, the lines $d_1$, $d_2$ and the points $B\in d_1$, $C\in d_2$ such that $d_1$, $d_2$ are medians in $\Delta ABC$, find $B$.

Question

Consider the point

$$A(0, -1)$$

and the lines:

$$d_1 : x-y+1=0$$

$$d_2 : 2x-y=0$$

and I am told that the point $B$ is on the line $d_1$, while the point $C$ is on the line $d_2$. Also, the lines $d_1$ and $d_2$ are median lines in the triangle $\Delta ABC$. Given all of this, I have to find the coordinates of $B$.

We can rewrite the two lines as:

$$d_1:y=x+1$$

$$d_2:y=2x$$

Since $B \in d_1$ and $C \in d_2$ that means that we have:

$$B(x_B, x_B+1)$$

$$C(x_C, 2x_C)$$

I created this drawing here to help me and I kept moving $B$ and $C$ along $d_1$ and $d_2$ but I don't see where should I place them such that $d_1$ and $d_2$ are median lines, and then how to find the coordinates of $B$.

Answer 1

Simply, mid-point of AB $(\frac{x_B}{2},\frac{x_B}{2})$ will lie on $d_2$ ie y=2x and you will get you $x_B=0$