𝐴(−2;3),𝐵(5;2)and𝐶(2;−1)areverticesof𝐴⃤𝐵𝐶inaCartesianplaneasshown below. θ is the angle of inclination of 𝐴𝐶.

Question

Figure 1: Diagram for question 1.
1.1
Prove that ⃤𝐴𝐵𝐶 is right-angled.
(5)
1.2
Calculatetheareaof𝐴⃤𝐵𝐶.
(5)
1.3
Calculate the size of θ to the nearest degree.
(3)
1.4
Prove the coordinates of the midpoint 𝑀 on line 𝐴𝐶 are (0; 1).
(3)
1.5
Hence, determine the equation of the line 𝑀𝑁 passing through 𝑀, which is parallel to 𝐶𝐵.
(4)
1.6
Determine whether the midpoint of 𝐴𝐵 lies on the line 𝑀𝑁.
(4)

Answer 1

1.1 To prove that ⃤𝐴𝐵𝐶 is right-angled, we can calculate the slopes of the two sides that form the right angle.

Slope of 𝐴𝐶 = (3 - (-1)) / (-2 - 2) = 4 / -4 = -1
Slope of 𝐵𝐶 = (2 - (-1)) / (5 - 2) = 3 / 3 = 1

Since the product of the slopes (-1 * 1) is -1, the two sides are perpendicular to each other, making ⃤𝐴𝐵𝐶 right-angled.

1.2 To calculate the area of ⃤𝐴𝐵𝐶, we can use the formula for the area of a triangle:
Area = 1/2 * base * height

First, we need to calculate the base and height of the triangle:
Base = Distance between point A and point B = √[(5 - (-2))^2 + (2 - 3)^2] = √[7^2 + (-1)^2] = √(49 + 1) = √50 = 5√2
Height = Distance between point C and line AB = Distance from C to line AB is the perpendicular distance from C to line AB. To calculate this, we can use the formula for the distance between a point and a line:
Distance = |ax1 + by1 + c| / √(a^2 + b^2)
In this case, the equation of line AB is determined by points A and B: y = -1/7x + 19/7
The coefficients a, b, and c in the equation ax + by + c = 0 are 1, 7, and -19, respectively. Substituting the coordinates of point C into the formula:
Distance = |1(2) + 7(-1) - 19| / √(1^2 + 7^2) = |2 - 7 - 19| / √(1 + 49) = |-24| / √50 = 24 / √50 = 24√50 / 50 = 12√50 / 25 = 12√2 / 5

Now, we can calculate the area:
Area = 1/2 * 5√2 * 12√2 / 5 = 30

Therefore, the area of ⃤𝐴𝐵𝐶 is 30 square units.

1.3 To calculate the size of θ to the nearest degree, we can use the tangent of the angle θ:
tan(θ) = (3 - (-1)) / (-2 - 2) = 4 / -4 = -1
θ = tan^(-1)(-1) = -45 degrees

Therefore, the size of θ to the nearest degree is 45 degrees.

1.4 To prove that the coordinates of the midpoint 𝑀 on line 𝐴𝐶 are (0; 1), we can calculate the average of the x-coordinates and y-coordinates of points A and C:
x-coordinate of 𝑀 = (−2 + 2) / 2 = 0
y-coordinate of 𝑀 = (3 + (−1)) / 2 = 1

Therefore, the coordinates of the midpoint 𝑀 on line 𝐴𝐶 are (0; 1).

1.5 Since the line 𝑀𝑁 passing through 𝑀 is parallel to 𝐶𝐵, the slope of line 𝐶𝐵 is equal to the slope of line 𝑀𝑁. The slope of line 𝐶𝐵 can be calculated as:
Slope of 𝐶𝐵 = (2 - (-1)) / (5 - 2) = 3 / 3 = 1

Therefore, the equation of the line 𝑀𝑁 passing through 𝑀 with slope 1 is:
y - 1 = 1(x - 0)
y = x + 1

So, the equation of the line 𝑀𝑁 passing through 𝑀, which is parallel to 𝐶𝐵, is y = x + 1.

1.6 To determine whether the midpoint of 𝐴𝐵 lies on the line 𝑀𝑁, we can substitute the coordinates of the midpoint 𝑀 (0, 1) into the equation of the line 𝑀𝑁:
1 = 0 + 1
1 = 1

Since the coordinates of the midpoint 𝑀 satisfy the equation of the line 𝑀𝑁, the midpoint of 𝐴𝐵 lies on the line 𝑀𝑁.