Chapter 1, Problem 11MCQ

a.

To determine

To find the coordinates of the library

Answer to Problem 11MCQ

(11,-5)

Explanation of Solution

Given information:

Map is drawn on the grid

The high school is at (3,1) and town hall is at (-5,7)

Formula used:

Mid-point (x, y)

  $x=\frac{x_1+x_2}{2},y=\frac{y_1+y_2}{2}$

Calculation: let the coordinates of high school (3, 1) be $x$ and $y$

And let the coordinates of town hall (-5, 7) be $x_1$ and $y_1$

Substituting values in the formula

  $3=\frac{-5+x_2}{2},1=\frac{7+y_2}{2}$

Using cross multiplication method

  $6=-5+x_2,2=7+y_2$

Solving it further

  $x_2=11,y_2=-5$

Therefore, the coordinates of the library .........(11, -5)

b.

To determine

To calculate the distance between school and hall

Answer to Problem 11MCQ

500 meters

Explanation of Solution

Given information:

Coordinates of school (3, 1) and town hall (-5, 7)

Formula used:

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

Calculation: let the coordinates of school (3, 1) be $x_1,y_1$ and town hall (-5, 7) be $x_2,y_2$

Substituting values in the formula

Distance = $\sqrt{{\left(-5-3\right)}^2+{\left(7-1\right)}^2}$

Solving it further

Distance = $\sqrt{{\left(-8\right)}^2+{\left(6\right)}^2}$

Solving further

Distance = $\sqrt{64+36}$

Calculating the distance

Distance = $\sqrt{100}=10$ units

It is given in the question that one unit is equal to fifty meters

Therefore, 10 units = $10\times 50$

Distance between school and hall is

500 meters