Question

line segment gh and point k are shown on the coordinate plane below. what is the distance from point k to the midpoint of $overline{gh}$, to the nearest tenth of a unit?

Explanation:

Step1: Find the mid - point of line segment \(GH\)

The mid - point formula for two points \((x_1,y_1)\) and \((x_2,y_2)\) is \((\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})\). For \(G(-5,8)\) and \(H(9,-2)\), \(x_1=-5,y_1 = 8,x_2=9,y_2=-2\).
The mid - point \(M\) has \(x\) - coordinate \(\frac{-5 + 9}{2}=\frac{4}{2}=2\) and \(y\) - coordinate \(\frac{8+( - 2)}{2}=\frac{6}{2}=3\). So the mid - point \(M\) of \(GH\) is \((2,3)\).

Step2: Use the distance formula

The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). Here, for \(K(-1,-3)\) and \(M(2,3)\), \(x_1=-1,y_1=-3,x_2 = 2,y_2=3\).
First, calculate \((x_2 - x_1)\) and \((y_2 - y_1)\): \(x_2 - x_1=2-(-1)=3\) and \(y_2 - y_1=3-(-3)=6\).
Then \(d=\sqrt{3^2+6^2}=\sqrt{9 + 36}=\sqrt{45}\approx6.7\)

Answer:

\(6.7\)