Chapter 4.8, Problem 33HP

To determine

Tocalculate:

The coordinates of point $L$ of isosceles triangle.

Answer to Problem 33HP

The coordinates of point $L$ is $\left(4a,0\right)$ .

Explanation of Solution

Given information:

Isosceles triangle.

Indicated triangle $=△JKL$

Coordinates:

  $\begin{array}{l}J\left(0,0\right)\\ K\left(2a,2b\right)\end{array}$

Formula Used:

Distance formula:

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

Calculation:

Since $△JKL$ is to be isosceles, choose a point for $L$ that is on the $x-$ axis such that $JL=KL$ .

The $y-$ coordinates of point $L$ will be $0$ . The $x-$ coordinates of $K$ must be halfway between $0$ and the $x-$ coordinate for $L$ because the triangle is isosceles.

Since, the $x-$ coordinate for $K$ is $2a$ , the $x-$ coordinate for $L$ must be $4a$ . So, place $L$ at $\left(4a,0\right)$ .

So, with the help of the distance formula it is easy to verify $JL=KL$ .

  

Now use the distance formula:

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

For $JK$ ,

Put the coordinates as $\left(x_{1,}{y}_1\right)=\left(0,0\right),\left(x_{2,}{y}_2\right)=\left(2a,2b\right)$

Simplify,

  $\begin{array}{l}JK=\sqrt{{\left(2a-0\right)}^2-{\left(2b-0\right)}^2}\\ =\sqrt{4a^2+4b^2}or2\sqrt{a^2+b^2}\end{array}$

For $KL$ ,

Put the coordinates as $\left(x_{1,}{y}_1\right)=\left(2a,2b\right),\left(x_{2,}{y}_2\right)=\left(4a,0\right)$

Simplify,

  $\begin{array}{l}KL=\sqrt{{\left(4a-2a\right)}^2-{\left(0-2b\right)}^2}\\ =\sqrt{4a^2+4b^2}or2\sqrt{a^2+b^2}\end{array}$

Since, $JL=KL,\overline{JL}\cong \overline{KL}$ and $△JKL$ is isosceles.

Therefore, the coordinates of point $L$ could be $\left(4a,0\right)$ .