$A (5, 1); B (1, 5)$ and $C (- 3, 1)$ are the vertices of $△ A B C$ , find the length of median $A D$ .

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Solution

In the given figure
$D$ is the mid-points of $B C$
Coordinates of $D = (\frac{1 + (- 3)}{2}, \frac{5 + (- 1)}{2})$
$∴ D = (- 1, 2)$
Now,
Length of $A D = \sqrt{{(Y_{2} - Y_{1})}^{2} + {(X_{2} - X_{1})}^{2}}$
$= \sqrt{{(1 - 2)}^{2} + {(5 - (- 1))}^{2}}$
$= \sqrt{{(- 1)}^{2} + {(6)}^{2}}$
$= \sqrt{37}$
Hence, the length of $A D$ is $\sqrt{37}$ sq. units.

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A(5, 1); B(1, 5) and C(−3, 1) are the vertices of △ ABC, find the length of median AD.

In the given figure D is the mid-points of BCCoordinates of D=(1+(−3)2,5+(−1)2)∴D=(−1,2)Now,Length of AD=√(Y2−Y1)2+(X2−X1)2=√(1−2)2+(5−(−1))2=√(−1)2+(6)2 =√37Hence, the length of AD is √37 sq. units.

$A (5, 1); B (1, 5)$ and $C (- 3, 1)$ are the vertices of $△ A B C$ , find the length of median $A D$ .