Chapter 1.3, Problem 32PPS

a.

To determine

Distance between houses.

Answer to Problem 32PPS

  $\sqrt{34}$ units is the distance.

Explanation of Solution

Given information:

Points marked as houses

Calculation:

Observing graph

P’s house point $(4,6)$

A’s house point $(7,1)$

To find the distance,

Let

  $x_1^{}=4,x_2^{}=7$ and

  $y_1^{}=6,y_2^{}=1$

Using distance formula,

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

Substituting values,

  $=\sqrt{{(7-4)}^2+{(1-6)}^2}$

Subtracting,

  $=\sqrt{{(3)}^2+{(-5)}^2}$

So,

  $=\sqrt{9+25}$

Hence, $\sqrt{34}$ units is the distance.

b.

To determine

To find the distance between

Answer to Problem 32PPS

The distance is $8\sqrt{2}$ .

Explanation of Solution

Given information:

Points marked as houses.

Calculation:

If P is moved 3 blocks to north,

New coordinates are $(4,9)$

And A is moved 5 blocks west,

New coordinates are $(12,1)$

Now to find the distance,

Let

  $x_1^{}=4,x_2^{}=12$ and,

  $y_1^{}=9,y_2^{}=1$

Using distance formula,

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

Substituting values,

Distance= $\sqrt{{(12-4)}^2+{(1-9)}^2}$

Solving,

Distance= $\sqrt{{(8)}^2+{(-8)}^2}$

Now,

Distance= $\sqrt{64+64}$

Adding,

Distance= $\sqrt{128}$

So now,

Distance= $8\sqrt{2}$

Hence, the distance is $8\sqrt{2}$ .