Question

3
what is the distance between
$(3,1)$ and $(-4,-3)$?

Explanation:

Step1: Find the differences in x - coordinates and y - coordinates

The formula for the distance \(d\) between two points \((x_1,y_1)\) and \((x_2,y_2)\) is based on the Pythagorean theorem \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).
Let \((x_1,y_1)=(3,1)\) and \((x_2,y_2)=(-4,-3)\).
The difference in \(x\) - coordinates: \(x_2 - x_1=-4 - 3=-7\).
The difference in \(y\) - coordinates: \(y_2 - y_1=-3 - 1=-4\).

Step2: Square the differences

\((x_2 - x_1)^2=(-7)^2 = 49\).
\((y_2 - y_1)^2=(-4)^2 = 16\).

Step3: Sum the squared differences

\((x_2 - x_1)^2+(y_2 - y_1)^2=49 + 16=65\).

Step4: Take the square root

\(d=\sqrt{65}\).

Answer:

\(a=-7\), \(b = - 4\), \(c=\sqrt{65}\)
\((-7)^2+(-4)^2 = 49 + 16\)
\(49+16 = 65\)
\(\sqrt{49 + 16}=\sqrt{65}\)
\(\sqrt{65}=\sqrt{65}\)