Question

q = (-4, -4)
r = (3, 3)

Explanation:

Step1: Identify the problem type

We need to find the distance between two points \( Q = (-4, -4) \) and \( R = (3, 3) \) in a coordinate plane. The distance formula between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is \( d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2} \).

Step2: Substitute the values into the formula

Here, \( x_1=-4,y_1 = - 4,x_2=3,y_2=3 \). Substitute these values into the distance formula:
\( d=\sqrt{(3 - (-4))^2+(3 - (-4))^2} \)
First, simplify the expressions inside the parentheses:
\( 3-(-4)=3 + 4=7 \) and \( 3-(-4)=7 \)
So the formula becomes \( d=\sqrt{7^2+7^2} \)

Step3: Calculate the squares and sum

Calculate \( 7^2 = 49 \), so \( 7^2+7^2=49 + 49=98 \)
Then \( d=\sqrt{98} \)

Step4: Simplify the square root

We know that \( 98=49\times2 \), so \( \sqrt{98}=\sqrt{49\times2}=\sqrt{49}\times\sqrt{2}=7\sqrt{2}\approx7\times1.414 = 9.899\) (if we want a decimal approximation) or we can leave it in radical form as \( 7\sqrt{2} \)

Answer:

The distance between \( Q(-4,-4) \) and \( R(3,3) \) is \( 7\sqrt{2}\) (or approximately \( 9.90 \))