Answer :
Let's break down the problems one by one and provide a clear, step-by-step solution for each of them.
### 1. Finding the Mid-point
To find the coordinates of the midpoint of the interval joining the points (1, 3) and (3, -1), we use the midpoint formula:
[tex]\[ \text{Midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \][/tex]
Given points are [tex]\((1, 3)\)[/tex] and [tex]\((3, -1)\)[/tex].
Here:
[tex]\( x_1 = 1 \)[/tex]
[tex]\( y_1 = 3 \)[/tex]
[tex]\( x_2 = 3 \)[/tex]
[tex]\( y_2 = -1 \)[/tex]
Plugging these into the formula, we get:
[tex]\[ \text{Midpoint} = \left( \frac{1 + 3}{2}, \frac{3 + (-1)}{2} \right) = \left( \frac{4}{2}, \frac{2}{2} \right) = (2.0, 1.0) \][/tex]
### 2. Isosceles Triangle Check
We need to show that the triangle with vertices A(1, 1), B(4, 5), and C(0, 2) is isosceles. To do this, we calculate the lengths of the sides AB, BC, and CA.
Distance formula:
[tex]\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Side AB:
[tex]\[ \text{AB} = \sqrt{(4 - 1)^2 + (5 - 1)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5.0 \][/tex]
Side BC:
[tex]\[ \text{BC} = \sqrt{(0 - 4)^2 + (2 - 5)^2} = \sqrt{(-4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5.0 \][/tex]
Side CA:
[tex]\[ \text{CA} = \sqrt{(1 - 0)^2 + (1 - 2)^2} = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \approx 1.4142135623730951 \][/tex]
Since two sides (AB and BC) are equal:
[tex]\[ AB = BC = 5.0 \][/tex]
The triangle is isosceles as it has at least two equal sides.
### 3. Right Triangle Verification
We need to prove that a right triangle is formed if the points P(1, 6), Q(-6, 10), and R(-3, 4) are joined. A triangle is right-angled if the dot product of vectors representing two sides is zero (indicating orthogonality).
Vectors:
[tex]\[ \overrightarrow{PQ} = (Q_x - P_x, Q_y - P_y) = (-6 - 1, 10 - 6) = (-7, 4) \][/tex]
[tex]\[ \overrightarrow{QR} = (R_x - Q_x, R_y - Q_y) = (-3 - (-6), 4 - 10) = (3, -6) \][/tex]
[tex]\[ \overrightarrow{RP} = (P_x - R_x, P_y - R_y) = (1 - (-3), 6 - 4) = (4, 2) \][/tex]
Dot products:
[tex]\[ \overrightarrow{PQ} \cdot \overrightarrow{QR} = (-7) \cdot 3 + 4 \cdot (-6) = -21 - 24 = -45 \][/tex]
[tex]\[ \overrightarrow{QR} \cdot \overrightarrow{RP} = 3 \cdot 4 + (-6) \cdot 2 = 12 - 12 = 0 \][/tex]
[tex]\[ \overrightarrow{RP} \cdot \overrightarrow{PQ} = 4 \cdot (-7) + 2 \cdot 4 = -28 + 8 = -20 \][/tex]
Since the dot product [tex]\(\overrightarrow{QR} \cdot \overrightarrow{RP} = 0\)[/tex], vectors QR and RP are orthogonal. Therefore, the angle between [tex]\(\overrightarrow{QR}\)[/tex] and [tex]\(\overrightarrow{RP}\)[/tex] is a right angle, proving that the triangle is a right triangle.
### Summary of Results
1. Midpoint of the interval joining points (1, 3) and (3, -1): (2.0, 1.0)
2. Lengths of sides of triangle A(1, 1), B(4, 5), C(0, 2): AB = 5.0, BC = 5.0, CA = 1.4142135623730951. The triangle is isosceles.
3. Triangle formed by points P(1, 6), Q(-6, 10), and R(-3, 4) is a right triangle.
### 1. Finding the Mid-point
To find the coordinates of the midpoint of the interval joining the points (1, 3) and (3, -1), we use the midpoint formula:
[tex]\[ \text{Midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \][/tex]
Given points are [tex]\((1, 3)\)[/tex] and [tex]\((3, -1)\)[/tex].
Here:
[tex]\( x_1 = 1 \)[/tex]
[tex]\( y_1 = 3 \)[/tex]
[tex]\( x_2 = 3 \)[/tex]
[tex]\( y_2 = -1 \)[/tex]
Plugging these into the formula, we get:
[tex]\[ \text{Midpoint} = \left( \frac{1 + 3}{2}, \frac{3 + (-1)}{2} \right) = \left( \frac{4}{2}, \frac{2}{2} \right) = (2.0, 1.0) \][/tex]
### 2. Isosceles Triangle Check
We need to show that the triangle with vertices A(1, 1), B(4, 5), and C(0, 2) is isosceles. To do this, we calculate the lengths of the sides AB, BC, and CA.
Distance formula:
[tex]\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Side AB:
[tex]\[ \text{AB} = \sqrt{(4 - 1)^2 + (5 - 1)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5.0 \][/tex]
Side BC:
[tex]\[ \text{BC} = \sqrt{(0 - 4)^2 + (2 - 5)^2} = \sqrt{(-4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5.0 \][/tex]
Side CA:
[tex]\[ \text{CA} = \sqrt{(1 - 0)^2 + (1 - 2)^2} = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \approx 1.4142135623730951 \][/tex]
Since two sides (AB and BC) are equal:
[tex]\[ AB = BC = 5.0 \][/tex]
The triangle is isosceles as it has at least two equal sides.
### 3. Right Triangle Verification
We need to prove that a right triangle is formed if the points P(1, 6), Q(-6, 10), and R(-3, 4) are joined. A triangle is right-angled if the dot product of vectors representing two sides is zero (indicating orthogonality).
Vectors:
[tex]\[ \overrightarrow{PQ} = (Q_x - P_x, Q_y - P_y) = (-6 - 1, 10 - 6) = (-7, 4) \][/tex]
[tex]\[ \overrightarrow{QR} = (R_x - Q_x, R_y - Q_y) = (-3 - (-6), 4 - 10) = (3, -6) \][/tex]
[tex]\[ \overrightarrow{RP} = (P_x - R_x, P_y - R_y) = (1 - (-3), 6 - 4) = (4, 2) \][/tex]
Dot products:
[tex]\[ \overrightarrow{PQ} \cdot \overrightarrow{QR} = (-7) \cdot 3 + 4 \cdot (-6) = -21 - 24 = -45 \][/tex]
[tex]\[ \overrightarrow{QR} \cdot \overrightarrow{RP} = 3 \cdot 4 + (-6) \cdot 2 = 12 - 12 = 0 \][/tex]
[tex]\[ \overrightarrow{RP} \cdot \overrightarrow{PQ} = 4 \cdot (-7) + 2 \cdot 4 = -28 + 8 = -20 \][/tex]
Since the dot product [tex]\(\overrightarrow{QR} \cdot \overrightarrow{RP} = 0\)[/tex], vectors QR and RP are orthogonal. Therefore, the angle between [tex]\(\overrightarrow{QR}\)[/tex] and [tex]\(\overrightarrow{RP}\)[/tex] is a right angle, proving that the triangle is a right triangle.
### Summary of Results
1. Midpoint of the interval joining points (1, 3) and (3, -1): (2.0, 1.0)
2. Lengths of sides of triangle A(1, 1), B(4, 5), C(0, 2): AB = 5.0, BC = 5.0, CA = 1.4142135623730951. The triangle is isosceles.
3. Triangle formed by points P(1, 6), Q(-6, 10), and R(-3, 4) is a right triangle.
