Question

given: g(-2, 8), h(-10, -8) find: gh

Explanation:

Step1: Recall distance formula

The distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). Here, \(G(-2,8)\) so \(x_1=-2,y_1 = 8\) and \(H(-10,-8)\) so \(x_2=-10,y_2=-8\).

Step2: Substitute values into formula

Calculate \(x_2 - x_1=-10-(-2)=-10 + 2=-8\) and \(y_2 - y_1=-8 - 8=-16\). Then, \((x_2 - x_1)^2=(-8)^2 = 64\) and \((y_2 - y_1)^2=(-16)^2 = 256\).

Step3: Sum the squares and take square root

Sum: \(64 + 256=320\). So \(GH=\sqrt{320}\). Simplify \(\sqrt{320}=\sqrt{64\times5}=8\sqrt{5}\)? Wait, no, wait: Wait, \((-8)^2=64\), \((-16)^2 = 256\), \(64 + 256=320\)? Wait, no, \(-10-(-2)=-8\), \(-8 - 8=-16\). Then \((-8)^2=64\), \((-16)^2 = 256\), \(64+256 = 320\)? Wait, but \(320=64\times5\)? No, \(64\times5 = 320\)? Wait, \(64\times5=320\), yes. But wait, another way: Wait, maybe I made a mistake. Wait, \(-10 - (-2)=-8\), correct. \(-8 - 8=-16\), correct. Then \((-8)^2=64\), \((-16)^2=256\), sum is \(64 + 256 = 320\). Then \(\sqrt{320}=\sqrt{64\times5}=8\sqrt{5}\)? Wait, but let's check again. Wait, \(8\sqrt{5}\) squared is \(64\times5 = 320\), yes. Wait, but wait, the options have \(8\sqrt{5}\) as one of them. Wait, but let's recalculate:

Wait, \(x_1=-2,y_1 = 8\); \(x_2=-10,y_2=-8\).

\(\Delta x=x_2 - x_1=-10-(-2)=-8\), \(\Delta y=y_2 - y_1=-8 - 8=-16\).

Then distance \(GH=\sqrt{(\Delta x)^2+(\Delta y)^2}=\sqrt{(-8)^2+(-16)^2}=\sqrt{64 + 256}=\sqrt{320}\).

Simplify \(\sqrt{320}\): \(320 = 64\times5\), so \(\sqrt{64\times5}=\sqrt{64}\times\sqrt{5}=8\sqrt{5}\).

Wait, but let's check the options. The options include \(8\sqrt{5}\), so that's the answer.

Wait, but wait, maybe I miscalculated \(\Delta y\). \(y_2 - y_1=-8 - 8=-16\), correct. \(\Delta x=-10 - (-2)=-8\), correct. Then squares: 64 and 256, sum 320, square root 8√5. Yes.

Answer:

\(8\sqrt{5}\)