Question

Q4
Find the length of median BD of Δ ABC, if A(7, -3), B(5, 3) and C(3, 1).

Answer 1

Answer

Step-by-step explanation:

A ( 7 , -3) & B( 5 , 3) & C ( 3 , 1 )

By using distance formula

AB = $\sqrt{(x_{1}- x_{2} )^{2} + (y_{1}- y_{2} )^{2} }$ =  $\sqrt{( 7 - 5)^{2} + ( -3 -3)^{2} }$ = $\sqrt{4 +36}$ = $\sqrt{40}$

AC = $\sqrt{(x_{1}- x_{2} )^{2} + (y_{1}- y_{2} )^{2} }$ = $\sqrt{( 7 - 3)^{2} + ( -3 -1)^{2} }$ =$\sqrt{16 +16}$ =$\sqrt{32}$

AD = $\frac{AC}{2}$             ( ∵ BD ⊥ AD)

AD = $\frac{\sqrt{32} }{2}$

Now by using pythagoras theoram in Δ ABD

BD = $\sqrt{AB^{2} - AD^{2} } = \sqrt{40 - \frac{32}{4} } = \frac{8\sqrt{2} }{2}$

Answer 2

Answer:

Therefore the length of median is 4 units.

Step-by-step explanation:

Given points A(7,-3) ,B(5,3) and C(3,1)

Therefore the mid point A and C is $(\frac{7+3}{2} ,\frac{-3+1}{2} )$=(5, -1)

Therefore the distance between of (5,-1) and (5,3) is

$\sqrt{(5-5)^2+(-1-3)^2}$=4 units

Therefore the length of median is 4 units.