# Two horses are ready to return to their barn after a long workout session at the track. the horses are at coordinates h(1,10)

###### Question:

## Answers

The given information are the coordinates of points H, Z and B. So, the first step to do is to plot these points on a Cartesian plane as shown in the attached picture. We can deduce visually that horse Z is closer to the barns than horse H. But, to further justify the answer, we have to provide the magnitude of the distance of horses H and Z to barn B. In this approach, we use the distance formula:

d = √(x₂ - x₁)² + (y₂ - y₁)²

Use the coordinates of the two points to know their linear distances. These are represented by the red and green lines for each of the horses.

Distance between H and B = √(⁻3 - 1)² + (⁻9-10)²

Distance between H and B = √(⁻3 - 1)² + (⁻9-10)²

Distance between H and B = 19.4165

Since the scale is 1 unit = 100 meters, the actual distance between horse H and barn B is 19.416*100 = 1,941.65 meters

Distance between Z and B = √(⁻3 - 10)² + (⁻9-1)²

Distance between Z and B = √(⁻3 - 10)² + (⁻9-1)²

Distance between Z and B = 16.4012

Since the scale is 1 unit = 100 meters, the actual distance between horse Z and barn B is 16.4012*100 = 1,640.12 meters

Comparing the distances: 1,941.65 meters > 1,640.12 meters. Therefore, it justifies that horse Z is nearer to the barn.

[tex]Two horses are ready to return to their barn after a long workout session at the track. the horses a[/tex]

[tex]\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &H({{ 1}}\quad ,&{{ 10}})\quad % (c,d) &B({{ -3}}\quad ,&{{ -9}}) \end{array}\quad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2} \\\\\\ HB=\sqrt{(-3-1)^2+(-9-10)^2}\implies HB=\sqrt{(-4)^2+(-19)^2} \\\\\\ HB=\sqrt{377}\implies HB\approx 19.4\cdot \stackrel{meters}{100}\implies HB\approx 1940~m\\\\ -------------------------------\\\\[/tex]

[tex]\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &Z({{ 10}}\quad ,&{{ 1}})\quad % (c,d) &B({{ -3}}\quad ,&{{ -9}}) \end{array}\quad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2} \\\\\\ ZB=\sqrt{(-3-10)^2+(-9-1)^2}\implies HB=\sqrt{(-13)^2+(-10)^2} \\\\\\ ZB=\sqrt{269}\implies ZB\approx 16.4\cdot \stackrel{meters}{100}\implies ZB\approx 1640~m[/tex]

[tex]\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &H({{ 1}}\quad ,&{{ 10}})\quad % (c,d) &B({{ -3}}\quad ,&{{ -9}}) \end{array}\quad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2} \\\\\\ HB=\sqrt{(-3-1)^2+(-9-10)^2}\implies HB=\sqrt{(-4)^2+(-19)^2} \\\\\\ HB=\sqrt{377}\implies HB\approx 19.4\cdot \stackrel{meters}{100}\implies HB\approx 1940~m\\\\ -------------------------------\\\\[/tex]

[tex]\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &Z({{ 10}}\quad ,&{{ 1}})\quad % (c,d) &B({{ -3}}\quad ,&{{ -9}}) \end{array}\quad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2} \\\\\\ ZB=\sqrt{(-3-10)^2+(-9-1)^2}\implies HB=\sqrt{(-13)^2+(-10)^2} \\\\\\ ZB=\sqrt{269}\implies ZB\approx 16.4\cdot \stackrel{meters}{100}\implies ZB\approx 1640~m[/tex]

Distance between points [tex]\left(x_1,y_1\right)[/tex] and [tex]\left( x_2,y_2 \right)[/tex] is

[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2[/tex]

Distance from H to B:

[tex]d=\sqrt{(1-(-3))^2+(10-(-9))^2}=\sqrt{16+361}=\sqrt{376}" /> units.

Distance from Z to B:

[tex]d=\sqrt{(10-(-3))^2+(1-(-9))^2}=\sqrt{169+100}=\sqrt{269}[/tex] units.

Horse Z is closer to the barn.

(The conversion to meters is not required; the question does not ask for actual distances, so "units" is OK.)

H(1,10) Z(10,1) B(-3,-9)

1st Calculate the distance H(horse) H to B(barn)

2nd Calculate the distance Z(horse) H to B(barn)

3rd Compare

1st distance H(horse) H to B(barn)

HB = √[(x₂-x₂)²+(y₂-y₁)₂] → √[-3-1)² + (-9-10)²]=√377 = 19.42

2nd distance ZB = √[(-3-10)²+(-9-1)²] = √269 = 16.4

HB in meter =19.42 x 100 = 1,942 m

ZB in meter = 16.4 x 100 = 1,640 m

Obviously HB > ZB and Z is closer to the Barn

Horse at Z (1640.1m from barn)

To answer this question you need to determine each horse distance to their barn. For horse H the distance would be

X: 1-(-3)= 4

Y: 10-(-9)= 19

If you use Pythagoras theory then the distance would be

BH^2= 4^2 + 19^2

BH^2= 377

BH= 19.416

Horse H is 1941.6m from the barn

For horse Z the distance would be:

X: 10- (-3)= 13

Y: 1-(-9)= 10

BZ^2 = 13^2 + 10^2

BZ^2 = 269

BZ= 16.401

Horse Z is 1640.1m from barn, a bit closer than horse H

The given coordinates are

Barn, B (-3,-9),

Horse H: (1, 10),

Horse Z: (10, 1).

Each unit on the coordinate plane represents 100 m.

Note that the distance between two coordinates (x₁, y₁) and (x₂, y₂) is

[tex]d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}[/tex]

The distance between horse H and the barn is

[tex]d_{H} = \sqrt{(1+3)^{2} + (10+9)^2} =19.4165[/tex]

That is, 1941.64 m = 1.94 km (nearest hundredth).

The distance between horse Z and the barn is

[tex]d_{Z} = \sqrt{(10+3)^{2} + (1+9)^{2}} =16.4012[/tex]

That is, 1640.12 m = 1.64 km (nearest hundredth)

Horse Z is closer to the barn.

The given information are the coordinates of points H, Z and B. So, the first step to do is to plot these points on a Cartesian plane as shown in the attached picture. We can deduce visually that horse Z is closer to the barns than horse H. But, to further justify the answer, we have to provide the magnitude of the distance of horses H and Z to barn B. In this approach, we use the distance formula:

d = √(x₂ - x₁)² + (y₂ - y₁)²

Use the coordinates of the two points to know their linear distances. These are represented by the red and green lines for each of the horses.

Distance between H and B = √(⁻3 - 1)² + (⁻9-10)²

Distance between H and B = √(⁻3 - 1)² + (⁻9-10)²

Distance between H and B = 19.4165

Since the scale is 1 unit = 100 meters, the actual distance between horse H and barn B is 19.416*100 = 1,941.65 meters

Distance between Z and B = √(⁻3 - 10)² + (⁻9-1)²

Distance between Z and B = √(⁻3 - 10)² + (⁻9-1)²

Distance between Z and B = 16.4012

Since the scale is 1 unit = 100 meters, the actual distance between horse Z and barn B is 16.4012*100 = 1,640.12 meters

Comparing the distances: 1,941.65 meters > 1,640.12 meters. Therefore, it justifies that horse Z is nearer to the barn.

[tex]Two horses are ready to return to their barn after a long workout session at the track. the horses a[/tex]

The given information are the coordinates of points H, Z and B. So, the first step to do is to plot these points on a Cartesian plane as shown in the attached picture. We can deduce visually that horse Z is closer to the barns than horse H. But, to further justify the answer, we have to provide the magnitude of the distance of horses H and Z to barn B. In this approach, we use the distance formula:

d = √(x₂ - x₁)² + (y₂ - y₁)²

Use the coordinates of the two points to know their linear distances. These are represented by the red and green lines for each of the horses.

Distance between H and B = √(⁻3 - 1)² + (⁻9-10)²

Distance between H and B = √(⁻3 - 1)² + (⁻9-10)²

Distance between H and B = 19.4165

Since the scale is 1 unit = 100 meters, the actual distance between horse H and barn B is 19.416*100 = 1,941.65 meters

Distance between Z and B = √(⁻3 - 10)² + (⁻9-1)²

Distance between Z and B = √(⁻3 - 10)² + (⁻9-1)²

Distance between Z and B = 16.4012

Since the scale is 1 unit = 100 meters, the actual distance between horse Z and barn B is 16.4012*100 = 1,640.12 meters

Comparing the distances: 1,941.65 meters > 1,640.12 meters. Therefore, it justifies that horse Z is nearer to the barn.

[tex]Two horses are ready to return to their barn after a long workout session at the track. the horses a[/tex]