Question

If A$(4,9)$, B$(2,3)$, and C$(6,5)$ are the vertices of ∆ABC, then the length of the median through C is
(a) $5$ units

(b) $\sqrt{10}$ units

(c) $\sqrt{25}$ units

(d) $10$ units

Answer 1

It is given that A$(4,9)$, B$(2,3)$, and C$(6,5)$ are the vertices of ∆ABC.

Let CD be the median of ∆ABC through C. Then, D is the mid-point of AB.
Using mid-point formula, coordinates of midpoint $=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$, we get

Here, Taking A$(x_1,y_1)$ and B$(x_2,y_2)$ points, $x_1=4$, $x_2=2,$ $y_1=9$, $y_2=3$
Coordinates of D $=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})=(\frac{4+2}{2},\frac{9+3}{2})$

$=(\frac{6}{2},\frac{12}{2})=(3,6)$

Using distance formula, Length between two points $=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

Here,Taking A$(x_1,y_1)$ and D$(x_2,y_2)$ points, $x_1=4$, $x_2=3$, $y_1=9$, $y_2=6$

Length of the median, AD $=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

$=\sqrt{(3-4{)}^2+(6-9{)}^2}$

$=\sqrt{(-1{)}^2+(-3{)}^2}$

$=\sqrt{10}$ units

Thus, the length of the required median is $\sqrt{10}$ units.
Hence, the correct answer is option B.