Question

4.
The triangle formed by the points (0,-1), (-2, 0), (-1,-3) is
a) an equilateral triangle
b) right angled isosceles triangle
c) scalene triangle
d) collinear
​

Answer 1

Answer:

Option (B): Right-angled isosceles triangle.

Step-by-step explanation:

To find the type of the triangle formed by the given points, we'll find the distances between the points taken two at a time, using the distance formula.

$\boxed{\sf Distance \ formula: \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}}$

For our convenience, let's name the points as A(0, -1), B(-2, 0), C(-1, -3).

Distance of AB:

x₁ → 0

x₂ → -2

y₁ → -1

y₂ → 0

$\sf AB = \sqrt{(0+2)^2+(-1-0)^2}\\ \\\sf AB = \sqrt{(2)^2+(-1)^2}\\ \\\sf AB = \sqrt{4+1}\\ \\\underline{\underline{\sf AB = \sqrt{5}}}$

Distance of BC:

x₁ → -2

x₂ → -1

y₁ → 0

y₂ → -3

$\sf BC = \sqrt{(-2-(-1))^2+(0-(-3))^2}\\ \\\sf BC = \sqrt{(-2+1)^2+(3)^2}\\ \\\sf BC = \sqrt{(-1)^2+9}\\ \\\sf BC = \sqrt{1+9}\\ \\\underline{\underline{\sf BC = \sqrt{10}}}$

Distance of AC:

x₁ → 0

x₂ → -1

y₁ → -1

y₂ → -3

$\sf AC = \sqrt{(0-(-1))^2+(-1-(-3))^2}\\ \\\sf AC = \sqrt{(1)^2+(-1+3)^2}\\ \\\sf AC = \sqrt{1+(2)^2}\\ \\\sf AC = \sqrt{1+4}\\ \\\underline{\underline{\sf AC = \sqrt{5}}}$

∴ AB = AC ≠ BC

Therefore the triangle can't be Equilateral, Scalene, or Collinear.

The remaining option is a right-angled isoceles triangle. If it is a right angled triangle, it should follow the pythaoras theorem.

$\boxed{\sf Pythagoras \ Theorem: Hypotenuse^2 = Altitude^2 + Base^2}$

In this triangle, BC is the longest side, therefore it is the hypotenuse.

$\Longrightarrow \sf BC^2 = AB^2 + AC^2\\ \\ \\\Longrightarrow \sf \left(\sqrt{10}\right)^2 = \left(\sqrt{5}\right)^2 + \left(\sqrt{5}\right)^2\\ \\ \\\Longrightarrow \sf 10 = 5 + 5\\ \\ \\\Longrightarrow \sf 10 = 10\\ \\ \\\sf LHS = RHS\\ \\\sf Hence \ proved.$

⇒ The ΔABC is a right angled isosceles triangle.