Chapter 4.8, Problem 2CYP

To determine

To find : missing coordinates of isosceles right triangle ABC.

Answer to Problem 2CYP

  $\begin{array}{l}A\left(0,a\right)\\ B\left(0,0\right)\\ C\left(a,0\right)\end{array}$

Explanation of Solution

Given information :

Calculation:

In above figure O represent the origin,

So, the coordinate of B is $\left(0,0\right)$ .

Now, the coordinate of C is on horizontal axis ( x-axis) and the points on the x-axis are always in the form of $\left(x,0\right)$ .

So, the coordinate of C is $\left(a,0\right)$ .

Now, find the distance between BC

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

Here, $\left(x_1,y_1\right)=\left(0,0\right)$ and $\left(x_2,y_2\right)=\left(a,0\right)$ .

Plug the values in distance formula

  $\begin{array}{l}=\sqrt{{\left(a-0\right)}^2+{\left(0-0\right)}^2}\\ =\sqrt{a^2}\\ =a\end{array}$

Here, $\Delta$ ABC is isosceles triangles and isosceles triangles have two equal side.

Here, Distance of AB is equal to distance of BC.

So, both the sides AB and AC are equal.

And the coordinate of A is on the vertical axis ( y-axis) and the points on the x-axis are always in the form of $\left(0,y\right)$ .

So, the coordinate of A is $\left(0,a\right)$ .