Chapter 1, Problem 48RE

To determine

To show:The given points are the vertices of an isosceles triangle .

Explanation of Solution

Given:

  $A=\left(3,4\right),B=\left(1,1\right)$ and $C=\left(-2,3\right)$

Formula Used:

The distance between two points :

  $d\left(P_1,P_2\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Proof:

First,find d (A,B).

For any points P₁(x₁, y₁) and P₂(x₂, y₂), the distance between the two points, denoted by d (P₁, P₂) is

  $d\left(P_1,P_2\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Substitute 1 for x₂, 3 for x₁, 1 for y₂, and 4 for y₁

  $d\left(A,B\right)=\sqrt{{\left(1-3\right)}^2+{\left(1-4\right)}^2}$

  $=\sqrt{4+9}$

  $=\sqrt{13}$

Similarly, use the distance formula to find d (B, C) and d (A, C).

  $d\left(B,C\right)=\sqrt{{\left(-2-1\right)}^2+{\left(3-1\right)}^2}$

  $=\sqrt{{\left(-3\right)}^2+2^2}$

  $=\sqrt{13}$

  $d\left(A,C\right)=\sqrt{{\left(-2-3\right)}^2+{\left(3-4\right)}^2}$

  $=\sqrt{{\left(-5\right)}^2+{\left(-1\right)}^2}$

  $=\sqrt{26}$

Now, check whether triangle is isosceles.

For a triangle to be isosceles, at least two of the sides must be of equal length.

Thus, $d\left(A,B\right)=d\left(B,C\right)=\sqrt{13}$

The triangle ABC is isosceles.

Conclusion:

Thus, triangle ABC is isosceles.