Question

triangle abc has coordinates a(-8,2), b(4,7), and c(2, -4). what is the length, to the nearest tenth, of bc? 1 of 11 question 6.3 11.2 7.6 13.0

Explanation:

Step1: Recall the distance formula

The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). For points \(B(4,7)\) and \(C(2,-4)\), \(x_1 = 4,y_1=7,x_2 = 2,y_2=-4\).

Step2: Substitute the values into the formula

First, find \((x_2 - x_1)\) and \((y_2 - y_1)\):
\(x_2 - x_1=2 - 4=-2\), \(y_2 - y_1=-4 - 7=-11\).
Then, \((x_2 - x_1)^2+(y_2 - y_1)^2=(-2)^2+(-11)^2=4 + 121=125\).

Step3: Calculate the distance

\(d=\sqrt{125}\approx11.2\) (since \(\sqrt{125}=\sqrt{25\times5}=5\sqrt{5}\approx5\times2.236 = 11.18\approx11.2\))

Answer:

11.2