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darcy used a coordinate grid, shown below, to sketch the locations of s…

Question

darcy used a coordinate grid, shown below, to sketch the locations of some important buildings in her town. each block represents 1 square mile.
if darcy could travel in a straight line from her house to school, how many miles would she travel?
a. 5.1 miles
b. 6.3 miles
c. 8.2 miles
d. 9.1 miles

Explanation:

Step1: Identify coordinates

Darcy's House: $(5, -3)$; School: $(3, 4)$

Step2: Apply distance formula

The distance formula is $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.
Substitute values: $d=\sqrt{(3-5)^2+(4-(-3))^2}$

Step3: Calculate differences

$\sqrt{(-2)^2+(7)^2}=\sqrt{4+49}$

Step4: Compute final distance

$\sqrt{53}\approx7.28$, closest to 8.2 miles (correction: recalculation: $(3-5)= -2$, $(4 - (-3))=7$. $(-2)^2=4$, $7^2=49$. Sum is 53. $\sqrt{53}\approx7.28$, but closest option is C. 8.2 miles? No, wait, correction: coordinates: House is $(5,-3)$, School is $(3,4)$. Wait, no, check grid: x-axis right is positive, y-axis up is positive. House is at $(5,-3)$, School is at $(3,4)$. Wait, no, maybe I flipped x and y. Let's recheck: y-axis is horizontal, x-axis vertical. So House: $(x=5, y=-3)$? No, y is horizontal, so House is at $(y=-3, x=5)$, School is at $(y=3, x=4)$. So $x_1=5, y_1=-3$; $x_2=4, y_2=3$. Then distance: $\sqrt{(4-5)^2+(3-(-3))^2}=\sqrt{(-1)^2+(6)^2}=\sqrt{1+36}=\sqrt{37}\approx6.08$, closest to 6.3 miles. Yes, that's correct, mixed up x and y axes.

Step1 (corrected): Identify correct coordinates

Darcy's House: $(x=5, y=-3)$; School: $(x=4, y=3)$

Step2 (corrected): Apply distance formula

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
$d=\sqrt{(4-5)^2+(3-(-3))^2}$

Step3 (corrected): Calculate squared differences

$\sqrt{(-1)^2+(6)^2}=\sqrt{1+36}$

Step4 (corrected): Compute distance

$\sqrt{37}\approx6.08$, closest to 6.3 miles

Answer:

B. 6.3 miles