1.1 To prove that ∆ABC is right-angled, we can calculate the slopes of the lines AB, BC, and AC. If any two of these are perpendicular (i.e., their slopes multiply to -1), then the triangle is right-angled.
Let's calculate the slopes:
Slope of AB = (2 - 3) / (5 + 2) = -1/7
Slope of BC = (-1 - 2) / (2 - 5) = -3/-3 = 1
Slope of AC = (-1 - 3) / (2 + 2) = -4/4 = -1
Since the slopes of AB and AC multiply to -1, the triangle ABC is right-angled.
1.2 To calculate the area of ∆ABC, we can use the formula for the area of a triangle with vertices A(x1, y1), B(x2, y2), and C(x3, y3) which is given by:
Area = 0.5 * |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|
Substitute the coordinates of points A, B, and C into the formula:
Area = 0.5 * |(-2)(2 - (-1)) + 5((-1) - 3) + 2(3 - 2)|
Area = 0.5 * |-2(-3) + 5(-4) + 2(1)|
Area = 0.5 * (6 - 20 + 2)
Area = 0.5 * (-12)
Area = -6
Therefore, the area of triangle ABC is 6 square units.
𝐴(−2;3),𝐵(5;2)and𝐶(2;−1)areverticesof𝐴⃤𝐵𝐶inaCartesianplaneasshown below. θ is the angle of inclination of 𝐴𝐶.
Figure 1: Diagram for question 1.
1.1
Prove that ⃤𝐴𝐵𝐶 is right-angled.
(5)
1.2
Calculatetheareaof𝐴⃤𝐵𝐶.
1 answer