First slide

Introduction to 3-D Geometry

Question

If $α, β$ are the roots of the quadratic equation $x^{2} - 5 x + 6 = 0$ such that $α < β$ and the vertices of a triangle are $A (3, 5, - 2), B (α, β, 5), C (- 1, β, α)$ ,then the triangle is

Easy

A

Isosceles
B

Equilateral
C

right angled
D

Scalene

Solution

The roots of the equation $x^{2} - 5 x + 6 = 0$ are $α = 2, β = 3$
The vertices of triangle are $A (3, 5, - 2), B (2, 3, 5), C (- 1, 3, 2)$
The distance between two points $A (x_{1}, y_{1}, z_{1})$ and $B (x_{2}, y_{2}, z_{2})$ is $A B = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2} + {(z_{2} - z_{1})}^{2}}$
Hence, the sides of the triangle are
$\begin{matrix} A B = \sqrt{1 + 4 + 49} = \sqrt{54} \\ B C = \sqrt{9 + 0 + 9} = \sqrt{18} \\ C A = \sqrt{16 + 4 + 16} = \sqrt{36} \end{matrix}$
Observing the lengths of sides of the triangle one is square of the sum of the squares of the other two
Hence the triangle is right angle triangle.

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