Chapter 4.8, Problem 31HP

To determine

Tocalculate:

The coordinates of point $L$ of scalene triangle.

Answer to Problem 31HP

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

Explanation of Solution

Given information:

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:

First choose the point on $x$ and $y$ axis with coordinates in terms of $a,b$ , such that no two sides of the triangle will be congruent.

Since point $J$ has coordinates $\left(0,0\right)$ and point $K$ has $\left(2a,2b\right)$ .

So, let point $L$ have coordinates $\left(a,0\right)$ , so use the distance formula to check that no two sides have the same length.

  

First use the distance formula:

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

So for $JK$ ,

  $\begin{array}{l}\left(x_{1,}{y}_1\right)=\left(0,0\right)\\ \left(x_{2,}{y}_2\right)=\left(2a,2b\right)\end{array}$

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 $JL$ ,

  $\begin{array}{l}\left(x_{1,}{y}_1\right)=\left(0,0\right)\\ \left(x_{2,}{y}_2\right)=\left(a,0\right)\end{array}$

Simplify,

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

For $KL$ ,

  $\begin{array}{l}\left(x_{1,}{y}_1\right)=\left(a,0\right)\\ \left(x_{2,}{y}_2\right)=\left(2a,2b\right)\end{array}$

Simplify,

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

Since, none of the sides have equal lengths the triangle is scalene.

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