Answer :
To find the vertices of the triangle ABC given that P(2, -1), Q(-2, 2), and R(-1, 4) are the midpoints of the sides BC, CA, and AB respectively, we can use the midpoint formula. The midpoint formula states that the midpoint M of a segment with endpoints [tex]\(X(x_1, y_1)\)[/tex] and [tex]\(Y(x_2, y_2)\)[/tex] is:
[tex]\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \][/tex]
Given the midpoints, we can set up equations to find the coordinates of the vertices A(x1, y1), B(x2, y2), and C(x3, y3).
### From midpoint P(2, -1):
P is the midpoint of BC:
[tex]\[ \left( \frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2} \right) = (2, -1) \][/tex]
This gives us two equations:
[tex]\[ \frac{x_2 + x_3}{2} = 2 \quad \Rightarrow \quad x_2 + x_3 = 4 \quad \text{(i)} \][/tex]
[tex]\[ \frac{y_2 + y_3}{2} = -1 \quad \Rightarrow \quad y_2 + y_3 = -2 \quad \text{(ii)} \][/tex]
### From midpoint Q(-2, 2):
Q is the midpoint of CA:
[tex]\[ \left( \frac{x_3 + x_1}{2}, \frac{y_3 + y_1}{2} \right) = (-2, 2) \][/tex]
This gives us two more equations:
[tex]\[ \frac{x_3 + x_1}{2} = -2 \quad \Rightarrow \quad x_3 + x_1 = -4 \quad \text{(iii)} \][/tex]
[tex]\[ \frac{y_3 + y_1}{2} = 2 \quad \Rightarrow \quad y_3 + y_1 = 4 \quad \text{(iv)} \][/tex]
### From midpoint R(-1, 4):
R is the midpoint of AB:
[tex]\[ \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = (-1, 4) \][/tex]
This gives us the final two equations:
[tex]\[ \frac{x_1 + x_2}{2} = -1 \quad \Rightarrow \quad x_1 + x_2 = -2 \quad \text{(v)} \][/tex]
[tex]\[ \frac{y_1 + y_2}{2} = 4 \quad \Rightarrow \quad y_1 + y_2 = 8 \quad \text{(vi)} \][/tex]
### Solving the System of Equations:
Now, we can solve this system of equations step-by-step.
1. Finding [tex]\(x_2\)[/tex]:
Add equations (i) and (v):
[tex]\[ x_2 + x_3 + x_1 + x_2 = 4 + (-2) \][/tex]
[tex]\[ 2x_2 + x_3 + x_1 = 2 \][/tex]
Using equation (iii) ([tex]\( x_3 + x_1 = -4 \)[/tex]):
[tex]\[ 2x_2 + (-4) = 2 \][/tex]
[tex]\[ 2x_2 = 6 \][/tex]
[tex]\[ x_2 = 3 \][/tex]
2. Finding [tex]\(x_3\)[/tex]:
Substitute [tex]\( x_2 = 3 \)[/tex] into equation (i):
[tex]\[ 3 + x_3 = 4 \][/tex]
[tex]\[ x_3 = 1 \][/tex]
3. Finding [tex]\(x_1\)[/tex]:
Substitute [tex]\( x_2 = 3 \)[/tex] into equation (v):
[tex]\[ x_1 + 3 = -2 \][/tex]
[tex]\[ x_1 = -5 \][/tex]
4. Finding [tex]\(y_2\)[/tex]:
Add equations (ii) and (vi):
[tex]\[ y_2 + y_3 + y_1 + y_2 = -2 + 8 \][/tex]
[tex]\[ 2y_2 + y_3 + y_1 = 6 \][/tex]
Using equation (iv) ([tex]\( y_3 + y_1 = 4 \)[/tex]):
[tex]\[ 2y_2 + 4 = 6 \][/tex]
[tex]\[ 2y_2 = 2 \][/tex]
[tex]\[ y_2 = 1 \][/tex]
5. Finding [tex]\(y_3\)[/tex]:
Substitute [tex]\( y_2 = 1 \)[/tex] into equation (ii):
[tex]\[ 1 + y_3 = -2 \][/tex]
[tex]\[ y_3 = -3 \][/tex]
6. Finding [tex]\(y_1\)[/tex]:
Substitute [tex]\( y_2 = 1 \)[/tex] into equation (vi):
[tex]\[ y_1 + 1 = 8 \][/tex]
[tex]\[ y_1 = 7 \][/tex]
### Final Vertices:
The vertices of the triangle ABC are:
[tex]\[ A = (-5, 7) \][/tex]
[tex]\[ B = (3, 1) \][/tex]
[tex]\[ C = (1, -3) \][/tex]
Therefore, the vertices of the triangle are [tex]\((-5, 7)\)[/tex], [tex]\((3, 1)\)[/tex], and [tex]\((1, -3)\)[/tex].
[tex]\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \][/tex]
Given the midpoints, we can set up equations to find the coordinates of the vertices A(x1, y1), B(x2, y2), and C(x3, y3).
### From midpoint P(2, -1):
P is the midpoint of BC:
[tex]\[ \left( \frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2} \right) = (2, -1) \][/tex]
This gives us two equations:
[tex]\[ \frac{x_2 + x_3}{2} = 2 \quad \Rightarrow \quad x_2 + x_3 = 4 \quad \text{(i)} \][/tex]
[tex]\[ \frac{y_2 + y_3}{2} = -1 \quad \Rightarrow \quad y_2 + y_3 = -2 \quad \text{(ii)} \][/tex]
### From midpoint Q(-2, 2):
Q is the midpoint of CA:
[tex]\[ \left( \frac{x_3 + x_1}{2}, \frac{y_3 + y_1}{2} \right) = (-2, 2) \][/tex]
This gives us two more equations:
[tex]\[ \frac{x_3 + x_1}{2} = -2 \quad \Rightarrow \quad x_3 + x_1 = -4 \quad \text{(iii)} \][/tex]
[tex]\[ \frac{y_3 + y_1}{2} = 2 \quad \Rightarrow \quad y_3 + y_1 = 4 \quad \text{(iv)} \][/tex]
### From midpoint R(-1, 4):
R is the midpoint of AB:
[tex]\[ \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = (-1, 4) \][/tex]
This gives us the final two equations:
[tex]\[ \frac{x_1 + x_2}{2} = -1 \quad \Rightarrow \quad x_1 + x_2 = -2 \quad \text{(v)} \][/tex]
[tex]\[ \frac{y_1 + y_2}{2} = 4 \quad \Rightarrow \quad y_1 + y_2 = 8 \quad \text{(vi)} \][/tex]
### Solving the System of Equations:
Now, we can solve this system of equations step-by-step.
1. Finding [tex]\(x_2\)[/tex]:
Add equations (i) and (v):
[tex]\[ x_2 + x_3 + x_1 + x_2 = 4 + (-2) \][/tex]
[tex]\[ 2x_2 + x_3 + x_1 = 2 \][/tex]
Using equation (iii) ([tex]\( x_3 + x_1 = -4 \)[/tex]):
[tex]\[ 2x_2 + (-4) = 2 \][/tex]
[tex]\[ 2x_2 = 6 \][/tex]
[tex]\[ x_2 = 3 \][/tex]
2. Finding [tex]\(x_3\)[/tex]:
Substitute [tex]\( x_2 = 3 \)[/tex] into equation (i):
[tex]\[ 3 + x_3 = 4 \][/tex]
[tex]\[ x_3 = 1 \][/tex]
3. Finding [tex]\(x_1\)[/tex]:
Substitute [tex]\( x_2 = 3 \)[/tex] into equation (v):
[tex]\[ x_1 + 3 = -2 \][/tex]
[tex]\[ x_1 = -5 \][/tex]
4. Finding [tex]\(y_2\)[/tex]:
Add equations (ii) and (vi):
[tex]\[ y_2 + y_3 + y_1 + y_2 = -2 + 8 \][/tex]
[tex]\[ 2y_2 + y_3 + y_1 = 6 \][/tex]
Using equation (iv) ([tex]\( y_3 + y_1 = 4 \)[/tex]):
[tex]\[ 2y_2 + 4 = 6 \][/tex]
[tex]\[ 2y_2 = 2 \][/tex]
[tex]\[ y_2 = 1 \][/tex]
5. Finding [tex]\(y_3\)[/tex]:
Substitute [tex]\( y_2 = 1 \)[/tex] into equation (ii):
[tex]\[ 1 + y_3 = -2 \][/tex]
[tex]\[ y_3 = -3 \][/tex]
6. Finding [tex]\(y_1\)[/tex]:
Substitute [tex]\( y_2 = 1 \)[/tex] into equation (vi):
[tex]\[ y_1 + 1 = 8 \][/tex]
[tex]\[ y_1 = 7 \][/tex]
### Final Vertices:
The vertices of the triangle ABC are:
[tex]\[ A = (-5, 7) \][/tex]
[tex]\[ B = (3, 1) \][/tex]
[tex]\[ C = (1, -3) \][/tex]
Therefore, the vertices of the triangle are [tex]\((-5, 7)\)[/tex], [tex]\((3, 1)\)[/tex], and [tex]\((1, -3)\)[/tex].
