Question

If the vertices of $\Delta$ABC are A(4, 2), B(6, 4), and C(4, 6), find the length of the median AD.

• A. $\sqrt{7}$ units
• B. $\sqrt{10}$ units
• C. $\sqrt{11}$ units
• D. $\sqrt{13}$ units

Answer 1

The correct option is B $\sqrt{10}$ units

Given: A(4, 2), B(6, 4), and C(4, 6)

AD is the median of $\Delta$ABC, so D is the midpoint of BC.

D = $(\frac{6+4}{2},\frac{4+6}{2})$
D = (5, 5)

AD = $\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$
AD = $\sqrt{(5-4{)}^2+(5-2{)}^2}$
AD = $\sqrt{(1{)}^2+(3{)}^2}$
​​​​​​​​​​​​​​AD = $\sqrt{1+9}$
​​​​​​​​​​​​​​​​​​​​​AD = $\sqrt{10}$ units