Question

In ABC, M is the midpoint of AB. Show that AM = MB = MC if the vertices ofAABC are A(7, 1), B(1, -7), and C(1, 1). Provide your complete solutions and proofsin your paper homework and enter the numeric answers online.

Answer 1

The vertices of triangle are given as,

$\begin{gathered} A(7,1) \\ B(1,-7) \\ C(1,1) \end{gathered}$

Also given that M is the midpoint of side AB,

$AM=MB$

The corresponding diagram is given below,

Consider that the coordinates of the midpoint of a line segment is the average of the coordinates of the endpoints.

So the coordinates of point M (x,y) will be,

$\begin{gathered} x=\frac{7+1}{2}=\frac{8}{2}=4 \\ y=\frac{1+(-7)}{2}=-3 \end{gathered}$

It is found that the coordinates of point M are (4,-3).

Consider the distance formula to obtain the distance between two points,

$d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}$

The distance between A (7,1) and M (4,-3) is calculated as,

$\begin{gathered} AM=\sqrt[]{(4_{}-7)^2+(-3-1)^2} \\ AM=\sqrt[]{(-3)^2+(-4)^2} \\ AM=\sqrt[]{9+16} \\ AM=\sqrt[]{25} \\ AM=5 \end{gathered}$

The distance between B (1,-7) and M (4,-3) is calculated as,

$\begin{gathered} BM=\sqrt[]{(4-1)^2+(-3-(-7))^2} \\ BM=\sqrt[]{(3)^2+(4)^2} \\ BM=\sqrt[]{9+16} \\ BM=\sqrt[]{25} \\ BM=5 \end{gathered}$

The distance between C (1,1) and M (4,-3) is calculated as,

$\begin{gathered} BM=\sqrt[]{(4-1)^2+(-3-(-7))^2} \\ BM=\sqrt[]{(3)^2+(4)^2} \\ BM=\sqrt[]{9+16} \\ BM=\sqrt[]{25} \\ BM=5 \end{gathered}$