Chapter 8.6, Problem 47E

Solution

To find: The type of $\Delta ABC$ that is $\Delta ABC$ is equilateral, isosceles or scalene.

  $\Delta ABC$ is a scalene triangle.

Given information:

The given points are $A\left(-1,2,4\right)$ , $B\left(0,6,-3\right)$ and $C\left(2,-4,1\right)$ .

Formula used:

Distance formula: The distance $d\left(P,Q\right)$ between the points $P\left(x_1,y_1,z_1\right)$ and $Q\left(x_2,y_2,z_2\right)$ is $d\left(P,Q\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2+{\left(z_2-z_1\right)}^2}$ .

Calculation:

Consider the vertices of $\Delta ABC$ as $A\left(-1,2,4\right)$ , $B\left(0,6,-3\right)$ and $C\left(2,-4,1\right)$ .

Then $AB$ , $BC$ and $AC$ will be the sides of $\Delta ABC$ .

Length of side $AB$

  $AB=d\left(A,B\right)$

Using the formula $d\left(P,Q\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2+{\left(z_2-z_1\right)}^2}$

Substitute $-1$ for $x_1$ , $2$ for $y_1$ , $4$ for $z_1$ , $0$ for $x_2$ , $6$ for $y_2$ and $-3$ for $z_2$ .

  $\begin{array}{c}AB=\sqrt{{\left(0-\left(-1\right)\right)}^2+{\left(6-2\right)}^2+{\left(-3-4\right)}^2}\\ AB=\sqrt{1+4^2+{\left(-7\right)}^2}\\ AB=\sqrt{1+16+49}\\ AB=\sqrt{66}\end{array}$

Length of side $BC$

  $BC=d\left(B,C\right)$

Using the formula $d\left(P,Q\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2+{\left(z_2-z_1\right)}^2}$

Substitute $0$ for $x_1$ , $6$ for $y_1$ , $-3$ for $z_1$ , $2$ for $x_2$ , $-4$ for $y_2$ and $1$ for $z_2$ .

  $\begin{array}{c}BC=\sqrt{{\left(2-0\right)}^2+{\left(-4-6\right)}^2+{\left(1-\left(-3\right)\right)}^2}\\ BC=\sqrt{2^2+{\left(-10\right)}^2+{\left(4\right)}^2}\\ BC=\sqrt{4+100+16}\\ BC=\sqrt{120}\end{array}$

Length of side $AC$

  $AC=d\left(A,C\right)$

Using the formula $d\left(P,Q\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2+{\left(z_2-z_1\right)}^2}$

Substitute $-1$ for $x_1$ , $2$ for $y_1$ , $4$ for $z_1$ , $2$ for $x_2$ , $-4$ for $y_2$ and $1$ for $z_2$ .

  $\begin{array}{c}AC=\sqrt{{\left(2-\left(-1\right)\right)}^2+{\left(-4-2\right)}^2+{\left(1-4\right)}^2}\\ AC=\sqrt{3^2+{\left(-6\right)}^2+{\left(-3\right)}^2}\\ AC=\sqrt{9+36+9}\\ AC=\sqrt{54}\end{array}$

From the above calculation it is clear that length of each side of $\Delta ABC$ is different.

Thus, $\Delta ABC$ is a scalene triangle.