Answer :
To determine the vertices of the triangle given the midpoints of its sides are [tex]\( (2,3) \)[/tex], [tex]\( (3,2) \)[/tex], and [tex]\( (4,3) \)[/tex], follow these steps:
1. Define the vertices of the triangle:
Let the vertices of the triangle be [tex]\( A(x_1, y_1) \)[/tex], [tex]\( B(x_2, y_2) \)[/tex], and [tex]\( C(x_3, y_3) \)[/tex].
2. Relate the midpoints to the vertices using the midpoint formula:
- The given midpoint [tex]\( M1(2,3) \)[/tex] is the midpoint of the side [tex]\( BC \)[/tex]. Therefore,
[tex]\[ \left( \frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2} \right) = (2,3) \][/tex]
From this, we get two equations:
\begin{align}
\frac{x_2 + x_3}{2} &= 2 \implies x_2 + x_3 = 4 \\
\frac{y_2 + y_3}{2} &= 3 \implies y_2 + y_3 = 6
\end{align}
- The given midpoint [tex]\( M2(3,2) \)[/tex] is the midpoint of the side [tex]\( AC \)[/tex]. Therefore,
[tex]\[ \left( \frac{x_1 + x_3}{2}, \frac{y_1 + y_3}{2} \right) = (3,2) \][/tex]
From this, we get two equations:
\begin{align}
\frac{x_1 + x_3}{2} &= 3 \implies x_1 + x_3 = 6 \\
\frac{y_1 + y_3}{2} &= 2 \implies y_1 + y_3 = 4
\end{align}
- The given midpoint [tex]\( M3(4,3) \)[/tex] is the midpoint of the side [tex]\( AB \)[/tex]. Therefore,
[tex]\[ \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = (4,3) \][/tex]
From this, we get two equations:
\begin{align}
\frac{x_1 + x_2}{2} &= 4 \implies x_1 + x_2 = 8 \\
\frac{y_1 + y_2}{2} &= 3 \implies y_1 + y_2 = 6
\end{align}
3. Solve these system of equations:
We have the system of equations:
\begin{align}
x_2 + x_3 &= 4 \\
y_2 + y_3 &= 6 \\
x_1 + x_3 &= 6 \\
y_1 + y_3 &= 4 \\
x_1 + x_2 &= 8 \\
y_1 + y_2 &= 6
\end{align}
4. Step-by-step solution:
4.1. Add the equations [tex]\( x_2 + x_3 = 4 \)[/tex] and [tex]\( x_1 + x_3 = 6 \)[/tex]:
[tex]\[ x_2 + x_3 + x_1 + x_3 = 4 + 6 \implies x_1 + x_2 + 2x_3 = 10 \][/tex]
Using [tex]\( x_1 + x_2 = 8 \)[/tex]:
[tex]\[ 8 + 2x_3 = 10 \implies 2x_3 = 2 \implies x_3 = 1 \][/tex]
4.2. Substitute [tex]\( x_3 = 1 \)[/tex] into [tex]\( x_1 + x_3 = 6 \)[/tex]:
[tex]\[ x_1 + 1 = 6 \implies x_1 = 5 \][/tex]
4.3. Substitute [tex]\( x_3 = 1 \)[/tex] into [tex]\( x_2 + x_3 = 4 \)[/tex]:
[tex]\[ x_2 + 1 = 4 \implies x_2 = 3 \][/tex]
4.4. Add the equations [tex]\( y_2 + y_3 = 6 \)[/tex] and [tex]\( y_1 + y_3 = 4 \)[/tex]:
[tex]\[ y_2 + y_3 + y_1 + y_3 = 6 + 4 \implies y_1 + y_2 + 2y_3 = 10 \][/tex]
Using [tex]\( y_1 + y_2 = 6 \)[/tex]:
[tex]\[ 6 + 2y_3 = 10 \implies 2y_3 = 4 \implies y_3 = 2 \][/tex]
4.5. Substitute [tex]\( y_3 = 2 \)[/tex] into [tex]\( y_1 + y_3 = 4 \)[/tex]:
[tex]\[ y_1 + 2 = 4 \implies y_1 = 2 \][/tex]
4.6. Substitute [tex]\( y_3 = 2 \)[/tex] into [tex]\( y_2 + y_3 = 6 \)[/tex]:
[tex]\[ y_2 + 2 = 6 \implies y_2 = 4 \][/tex]
5. Final vertices of the triangle:
[tex]\[ A(x_1, y_1) = (5, 2) \][/tex]
[tex]\[ B(x_2, y_2) = (3, 4) \][/tex]
[tex]\[ C(x_3, y_3) = (1, 2) \][/tex]
Thus, the vertices of the triangle are [tex]\( (5, 2) \)[/tex], [tex]\( (3, 4) \)[/tex], and [tex]\( (1, 2) \)[/tex].
1. Define the vertices of the triangle:
Let the vertices of the triangle be [tex]\( A(x_1, y_1) \)[/tex], [tex]\( B(x_2, y_2) \)[/tex], and [tex]\( C(x_3, y_3) \)[/tex].
2. Relate the midpoints to the vertices using the midpoint formula:
- The given midpoint [tex]\( M1(2,3) \)[/tex] is the midpoint of the side [tex]\( BC \)[/tex]. Therefore,
[tex]\[ \left( \frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2} \right) = (2,3) \][/tex]
From this, we get two equations:
\begin{align}
\frac{x_2 + x_3}{2} &= 2 \implies x_2 + x_3 = 4 \\
\frac{y_2 + y_3}{2} &= 3 \implies y_2 + y_3 = 6
\end{align}
- The given midpoint [tex]\( M2(3,2) \)[/tex] is the midpoint of the side [tex]\( AC \)[/tex]. Therefore,
[tex]\[ \left( \frac{x_1 + x_3}{2}, \frac{y_1 + y_3}{2} \right) = (3,2) \][/tex]
From this, we get two equations:
\begin{align}
\frac{x_1 + x_3}{2} &= 3 \implies x_1 + x_3 = 6 \\
\frac{y_1 + y_3}{2} &= 2 \implies y_1 + y_3 = 4
\end{align}
- The given midpoint [tex]\( M3(4,3) \)[/tex] is the midpoint of the side [tex]\( AB \)[/tex]. Therefore,
[tex]\[ \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = (4,3) \][/tex]
From this, we get two equations:
\begin{align}
\frac{x_1 + x_2}{2} &= 4 \implies x_1 + x_2 = 8 \\
\frac{y_1 + y_2}{2} &= 3 \implies y_1 + y_2 = 6
\end{align}
3. Solve these system of equations:
We have the system of equations:
\begin{align}
x_2 + x_3 &= 4 \\
y_2 + y_3 &= 6 \\
x_1 + x_3 &= 6 \\
y_1 + y_3 &= 4 \\
x_1 + x_2 &= 8 \\
y_1 + y_2 &= 6
\end{align}
4. Step-by-step solution:
4.1. Add the equations [tex]\( x_2 + x_3 = 4 \)[/tex] and [tex]\( x_1 + x_3 = 6 \)[/tex]:
[tex]\[ x_2 + x_3 + x_1 + x_3 = 4 + 6 \implies x_1 + x_2 + 2x_3 = 10 \][/tex]
Using [tex]\( x_1 + x_2 = 8 \)[/tex]:
[tex]\[ 8 + 2x_3 = 10 \implies 2x_3 = 2 \implies x_3 = 1 \][/tex]
4.2. Substitute [tex]\( x_3 = 1 \)[/tex] into [tex]\( x_1 + x_3 = 6 \)[/tex]:
[tex]\[ x_1 + 1 = 6 \implies x_1 = 5 \][/tex]
4.3. Substitute [tex]\( x_3 = 1 \)[/tex] into [tex]\( x_2 + x_3 = 4 \)[/tex]:
[tex]\[ x_2 + 1 = 4 \implies x_2 = 3 \][/tex]
4.4. Add the equations [tex]\( y_2 + y_3 = 6 \)[/tex] and [tex]\( y_1 + y_3 = 4 \)[/tex]:
[tex]\[ y_2 + y_3 + y_1 + y_3 = 6 + 4 \implies y_1 + y_2 + 2y_3 = 10 \][/tex]
Using [tex]\( y_1 + y_2 = 6 \)[/tex]:
[tex]\[ 6 + 2y_3 = 10 \implies 2y_3 = 4 \implies y_3 = 2 \][/tex]
4.5. Substitute [tex]\( y_3 = 2 \)[/tex] into [tex]\( y_1 + y_3 = 4 \)[/tex]:
[tex]\[ y_1 + 2 = 4 \implies y_1 = 2 \][/tex]
4.6. Substitute [tex]\( y_3 = 2 \)[/tex] into [tex]\( y_2 + y_3 = 6 \)[/tex]:
[tex]\[ y_2 + 2 = 6 \implies y_2 = 4 \][/tex]
5. Final vertices of the triangle:
[tex]\[ A(x_1, y_1) = (5, 2) \][/tex]
[tex]\[ B(x_2, y_2) = (3, 4) \][/tex]
[tex]\[ C(x_3, y_3) = (1, 2) \][/tex]
Thus, the vertices of the triangle are [tex]\( (5, 2) \)[/tex], [tex]\( (3, 4) \)[/tex], and [tex]\( (1, 2) \)[/tex].
