Question

The lengths of the medians of $△$ ABC whose vertices are $A(7,-3),B(5,3)$ and $C(3,-1)$ are

• A. Medians: $AD=5$ units, $BE=5$ units & $CF=\sqrt{5}$ units
• B. Medians: $AD=5$ units, $BE=\sqrt{10}$ units & $CF=\sqrt{10}$ units
• C. Medians $AD=5$ units, $BE=5$ units & $CF=\sqrt{10}$ units
• D. Medians $AD=10$ units, $BE=\sqrt{5}$ units & $CF=\sqrt{10}$ units

Answer 1

Answer: (C)

Medians $AD=5$ units, $BE=5$ units & $CF=\sqrt{10}$ units

Co-ordinate of D :

$\because D$ is a midpoint of $BC$

$\therefore D_x=\frac{C_x+B_x}{2}=\frac{3+5}{2}=4$

$\therefore D_y=\frac{C_y+B_y}{2}=\frac{-1+3}{2}=1$

$\Rightarrow D\equiv (4,1)$

Co-ordinate of E :

$\because E$ is a midpoint of $AC$

$\therefore E_x=\frac{A_x+C_x}{2}=\frac{7+3}{2}=5$

$\therefore E_y=\frac{A_y+C_y}{2}=\frac{-3-1}{2}=-2$

$\Rightarrow E\equiv (5,-2)$

Co-ordinate of F :

$\because F$ is a midpoint of $AB$

$\therefore F_x=\frac{A_x+B_x}{2}=\frac{7+5}{2}=6$

$\therefore F_y=\frac{A_y+B_y}{2}=\frac{-3+3}{2}=0$

$\Rightarrow F\equiv (6,0)$

Length Of medians can be obtained Using Distance Formula

$d=\sqrt{(x_2-x_1{)}^2+(y_2-y1{)}^2}$

$\therefore AD=\sqrt{(4-7{)}^2+(1-(-3){)}^2}$

$\therefore AD=\sqrt{9+16}$

$\therefore AD=5$

$\therefore BE=\sqrt{(5-5{)}^2+(-2-3{)}^2}$

$\therefore BE=\sqrt{0+25}$

$\therefore BE=5$

$\therefore CF=\sqrt{(6-3{)}^2+(0-(-1){)}^2}$

$\therefore CF=\sqrt{9+1}$

$\therefore CF=\sqrt{10}$

Hence, option $C$ is correct.