Question

$A(5,1)$, $B(1,5)$ and $C(-3,-1)$ are the vertices of $\Delta ABC$. The length of its median AD is:

• A. $\sqrt{34}$
• B. $\sqrt{35}$
• C. $\sqrt{37}$
• D. $6$

Answer 1

Answer: (C)

$\sqrt{37}$

A median of a triangle is a line segment that joins the vertex of a triangle to the midpoint of the opposite side.
Mid point of two points ${(x}_1,y_1)$ and ${(x}_2,y_2)$ is calculated by the formula $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$
Since AD is the median, this means, D is the mid point of BC.

Using this formula,
mid point of BC $=(\frac{1-3}{2},\frac{5-1}{2})=(-1,2)$

Distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ can be calculated
using the formula $\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$
Distance between the points A$(5,1)$ and D $(-1,2)=\sqrt{{(-1-5)}^2+{(2-1)}^2}=\sqrt{36+1}=\sqrt{37}$