Question

the midpoints of a triangle is (1,5,-1) (0,4,-2) and (2,3,4) find its vertices

Answer 1

Question #27620

The midpoints of a triangle is (1,5,-1) (0,4,-2) and (2,3,4) find its vertices

Solution. Let $ABC$ be a triangle and $M(1,5,-1), N(0,4,-2)$ and $K(2,3,4)$ be the midpoints of the sides $AB, BC$ and $AC$, respectively. Denote the coordinates of the vertices by $A(x_{1},y_{1},z_{1}), B(x_{2},y_{2},z_{2}), C(x_{3},y_{3},z_{3})$. Then, using the formula for the coordinates of a midpoint, we obtain:

$$\left\{ \begin{array}{l} \frac {x _ {1} + x _ {2}}{2} = 1 \\ \frac {x _ {1} + x _ {3}}{2} = 2 \\ \frac {x _ {2} + x _ {3}}{2} = 0 \end{array} \right.$$

It follows that $x_{2} = -x_{3}$ and $x_{1} = 4 - x_{3}$. The first equation implies that $4 - 2x_{3} = 2$ and so $x_{3} = 1$.

Then $x_{2} = -1, x_{1} = 3$.

By analogy we have

$$\left\{ \begin{array}{l} \frac {y _ {1} + y _ {2}}{2} = 5 \\ \frac {y _ {1} + y _ {3}}{2} = 3 \\ \frac {y _ {2} + y _ {3}}{2} = 4 \end{array} \right.$$

Then $y_{2} = 10 - y_{1}, y_{3} = 6 - y_{1}$. It follows from the third equation that $y_{1} = 4$ and so $y_{2} = 6, y_{3} = 2$.

By the same method, we obtain

$$\left\{ \begin{array}{l} \frac {z _ {1} + z _ {2}}{2} = - 1 \\ \frac {z _ {1} + z _ {3}}{2} = 4 \\ \frac {z _ {2} + z _ {3}}{2} = - 2 \end{array} \right.$$

Thus $z_{2} = -2 - z_{1}$ and $z_{3} = 8 - z_{1}$. The third equation implies that $2z_{1} = 10, z_{1} = 5$. Thus $z_{2} = -7$, $z_{3} = 3$.

Finally, $A(3,4,5), B(-1,6,-7), C(1,2,3)$.

Answer. (3,4,5), $(-1,6,-7)$ (1,2,3).