Chapter 6.2, Problem 30PPS

a.

To determine

To calculate:

The diagonals halfway of the given figure if diagonals of $\mathrm{J}\mathrm{K}\mathrm{L}\mathrm{M}$ bisect each other.

Answer to Problem 30PPS

The diagonals halfway of the given figure if diagonals of $\mathrm{J}\mathrm{K}\mathrm{L}\mathrm{M}$ bisect each other are calculated as:

  $\begin{array}{l}\mathrm{J}\mathrm{P}=\mathrm{L}\mathrm{P}=\sqrt{13}\\ \mathrm{M}\mathrm{P}=\mathrm{K}\mathrm{P}=\sqrt{34}\end{array}$

Explanation of Solution

Given information:

  

Formula used:

The distance formula is given by the following expression:

  $\mathrm{d}=\sqrt{{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2+{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}^2}$

Calculation:

From the given graph of the parallelogram with the vertices, understands that:

  $\mathrm{K}\left(2,5\right),\mathrm{L}\left(-1,-1\right),\mathrm{K}\left(-8,-1\right),\mathrm{J}\left(-5,5\right),\mathrm{P}\left(-3,2\right)$

The diagonal halfway lengths are calculated by using the distance formula as:

  $\begin{array}{l}\mathrm{J}\mathrm{P}=\sqrt{{\left(-3+5\right)}^2+{\left(2-5\right)}^2}=\sqrt{13}\\ \mathrm{L}\mathrm{P}=\sqrt{{\left(-3+1\right)}^2+{\left(2+1\right)}^2}=\sqrt{13}\\ \mathrm{M}\mathrm{P}=\sqrt{{\left(-3+8\right)}^2+{\left(2+1\right)}^2}=\sqrt{34}\\ \mathrm{K}\mathrm{P}=\sqrt{{\left(-3-2\right)}^2+{\left(2-5\right)}^2}=\sqrt{34}\end{array}$

From the above calculations:

  $\begin{array}{l}\mathrm{J}\mathrm{P}=\mathrm{L}\mathrm{P}=\sqrt{13}\\ \mathrm{M}\mathrm{P}=\mathrm{K}\mathrm{P}=\sqrt{34}\end{array}$

(Since diagonals bisect each other)

b.

To determine

To explain:

The diagonals of $▱\mathrm{J}\mathrm{K}\mathrm{L}\mathrm{M}$ are congruent or not.

Answer to Problem 30PPS

The diagonals of are not congruent since diagonals of $▱\mathrm{J}\mathrm{K}\mathrm{L}\mathrm{M}$ are not of same measure.

Explanation of Solution

Given information:

  

From sub-part (a):

The diagonals of $▱\mathrm{J}\mathrm{K}\mathrm{L}\mathrm{M}$ are calculated as:

  $\begin{array}{l}\mathrm{J}\mathrm{P}=\mathrm{L}\mathrm{P}=\sqrt{13}\\ \mathrm{M}\mathrm{P}=\mathrm{K}\mathrm{P}=\sqrt{34}\end{array}$

Know that:

If the diagonals of the parallelogram are of same measure only then it can be said that it’s The diagonals are congruent in nature.

Here, two diagonals $\mathrm{J}\mathrm{L}$ and $\mathrm{K}\mathrm{M}$ are not same because:

  $\begin{array}{l}\mathrm{J}\mathrm{L}=\mathrm{J}\mathrm{P}+\mathrm{L}\mathrm{P}=\sqrt{13}+\sqrt{13}=2\sqrt{13}\\ \mathrm{K}\mathrm{M}=\mathrm{M}\mathrm{P}+\mathrm{K}\mathrm{P}=\sqrt{34}+\sqrt{34}=2\sqrt{34}\end{array}$

So,

  $\mathrm{J}\mathrm{L}\ne \mathrm{M}\mathrm{K}$

Hence, the diagonals of $▱\mathrm{J}\mathrm{K}\mathrm{L}\mathrm{M}$ are not congruent .

c.

To determine

To explain:

The slopes of two perpendicular lines are negative reciprocals of each other or not.

Answer to Problem 30PPS

The slopes of two perpendicular lines are not negative reciprocals of each other.

Explanation of Solution

Given information:

  

From the given graph of the parallelogram with the vertices, understands that:

  $\mathrm{K}\left(2,5\right),\mathrm{L}\left(-1,-1\right),\mathrm{K}\left(-8,-1\right),\mathrm{J}\left(-5,5\right),\mathrm{P}\left(-3,2\right)$

Know that:

The slope formula is given by the following expression:

  $\mathrm{m}=\frac{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}$

Now, find the slope values of two perpendicular lines $\overline{\mathrm{J}\mathrm{K}}$ and $\overline{\mathrm{J}\mathrm{M}}$ as:

  $\begin{array}{l}\overline{\mathrm{J}\mathrm{K}}=\frac{5-5}{2+5}=0\\ \overline{\mathrm{J}\mathrm{M}}=\frac{-1-5}{-8+5}=2\end{array}$

Here, slopes are $\overline{\mathrm{J}\mathrm{K}}=0$ and $\overline{\mathrm{J}\mathrm{M}}=2$ .

These are not negative reciprocals of each other.