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question find the distance between the two points in simplest radical f…

Question

question
find the distance between the two points in simplest radical form.
$(-1,-1)$ and $(5,-9)$

Explanation:

Step1: Identify the coordinates

Let \((x_1,y_1)=(-1,-1)\) and \((x_2,y_2)=(5,-9)\)

Step2: Use the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)

Substitute the values: \(d=\sqrt{(5-(-1))^2+(-9 - (-1))^2}\)

Step3: Simplify the expressions inside the square root

\(5-(-1)=5 + 1=6\) and \(-9-(-1)=-9 + 1=-8\)
So \(d=\sqrt{6^2+(-8)^2}\)

Step4: Calculate the squares

\(6^2 = 36\) and \((-8)^2=64\)
Then \(d=\sqrt{36 + 64}\)

Step5: Add the numbers inside the square root

\(36+64 = 100\)
So \(d=\sqrt{100}\)
But wait, let's check again. Wait, no, we made a mistake above. Wait the formula is correct. Wait another way:

$$ LATEXBLOCK0 $$

Wait no, wait the problem says "simplest radical form". But \(\sqrt{100}=10\) which is an integer. But let's re - derive properly.

The distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)

Here \(x_1=-1,y_1=-1,x_2 = 5,y_2=-9\)

\(x_2-x_1=5-(-1)=6\), \(y_2 - y_1=-9-(-1)=-8\)

\(d=\sqrt{6^{2}+(-8)^{2}}=\sqrt{36 + 64}=\sqrt{100}=10\)

Answer:

\(10\)