Triangles: Notes and Questions

Q. Find the incentre of the triangle the coordinates of whose vertices are given by A(x₁, y₁), B(x₂, y₂), C(x₃, y₃).

Solution:

By geometry, we know that BD/DC = AB/AC (since AD bisects ÐA).
The lengths of the sides AB, BC and AC are c, a and b respectively, then BD/DC = AB/AC = c/b.
Coordinates of D are (bx₂+cx₃/b+c, by₂+cy₃/b+c)
IB bisects DB. Hence ID/IA = BD/BA = (ac/b+c)/c = a/c+b.
Let the coordinates of I be (x, y).
Then x = ax₁+bx₂+cx₃/a+b+c, y = ay₁+by₂+cy₃/a+b+c.

Q. If (0, 1), (1, 1) and (1, 0) are middle points of the sides of a triangle, find its incentre.

Solution:

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃)be teh vertices of a triangle.
x₁ + x₂ = 0, x₂ + x₃ = 0, x₃ + x₁ = 0
y₁ + y₂ = 0, y₂ + y₃ = 0, y₃ + y₁ = 0.
Solving these equations, we get A(0, 0), B(0, 2) and C(2, 0).
Now, a = BC = 2√ 2, b = CA = 2 and c = AB = 2.
Thus, incentre of the triangle ABC is (2-√ 2, 2-√ 2).

Q. If midpoints of the sides of a triangle are (0, 4), (6, 4) and (6, 0), then find the vertices of triangle, centroid and circumcentre of triangle.

Solution:

Let points A (x₁, y₁), B (x₂, y₂) and C (x₃, y₃) be vertices of ΔABC.
x₁ + x₃ = 0 , y₁ + y₃ = 8
x₂ + x₃ = 12 , y₂ + y₃ = 8
x₁ + x₂ = 12, y₁ + y₂ = 0
Solving we get A (0, 0), B (12, 0) and C (0, 8)
Hence ΔABC is right angled triangle. ∠A = π/2
Circumcentre is midpoint of hypotenuse which is (6, 4) itself and centroid
(x₁+x₂+x₃)/3 , (y₁+y₂+y₃)/3 = (4 , 8/3)