Question

find the distance between the two points rounding to the nearest tenth (if necessary). (1,4) and (-2,8)

Explanation:

Step1: Recall the distance formula

The distance \( d \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by \( d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2} \). Here, \( x_1 = 1 \), \( y_1 = 4 \), \( x_2=-2 \), \( y_2 = 8 \).

Step2: Substitute the values into the formula

First, calculate \( x_2 - x_1=-2 - 1=-3 \) and \( y_2 - y_1=8 - 4 = 4 \). Then, substitute these into the formula: \( d=\sqrt{(-3)^2+4^2} \).

Step3: Simplify the expression inside the square root

Calculate \( (-3)^2 = 9 \) and \( 4^2 = 16 \). Then, \( 9 + 16=25 \). So, \( d=\sqrt{25} \). But wait, that's a mistake. Wait, no: \( (-3)^2=9 \), \( 4^2 = 16 \), \( 9 + 16 = 25 \)? Wait, no, \( 9+16 = 25 \), but \( \sqrt{25}=5 \)? Wait, no, wait: \( (x_2 - x_1)=-2 - 1=-3 \), \( (y_2 - y_1)=8 - 4 = 4 \). Then \( (-3)^2=9 \), \( 4^2 = 16 \), sum is \( 9 + 16 = 25 \), square root of 25 is 5. Wait, but let me check again. Wait, \( (x_1,y_1)=(1,4) \), \( (x_2,y_2)=(-2,8) \). So \( x_2 - x_1=-2 - 1=-3 \), \( y_2 - y_1=8 - 4 = 4 \). Then \( (x_2 - x_1)^2=(-3)^2 = 9 \), \( (y_2 - y_1)^2=4^2 = 16 \). Sum is \( 9 + 16 = 25 \), square root of 25 is 5. So the distance is 5.0 (since we can write it to the nearest tenth as 5.0).

Answer:

5.0