Question

Given A(5,1) B(1,5) C(-3, -1) are the vertices of the triangle ABC find the length of the median AD

Answer 1

Answer:

length of AD = $\sqrt{37}$

Step-by-step explanation:

Given : ΔABC with vertices A(5,1) , B(1,5) , C(-3,-1)

           AD is median.

To Find : Length of AD

Solution :

Since AD is median . So, it divides line BC into two equal parts i.e.

BD : DC = 1:1

Thus D is the midpoint of BC

Now to find points of D  

We will use mid point formula:

$(x,y) = (\frac{x_{1} +x_{2}}{2},\frac{y_{1} +y_{2}}{2})$

Since we are given that :

Coordinates of D = (x,y)

Coordinates of B = $(x_{1} ,y_{1} )$ = (1,5)

Coordinates of C = $(x_{2} ,y_{2} )$ = (-3,-1)

Putting values in Formula:

⇒$(x,y) = (\frac{1 -3}{2},\frac{5 -1}{2} )$

⇒$(x,y) = (\frac{-2}{2},\frac{4}{2} )$

⇒$(x,y) = (-1,2)$

Thus Coordinates of D = (-1,2)

Now to find length of AD , use distance formula:

$d=\sqrt{(x_{2} -x_{1}) ^{2}+(y_{2} -y_{1}) ^{2} }$

Using coordinates of A and D

$d=\sqrt{(-1 -5) ^{2}+(2 -1) ^{2} }$

$d=\sqrt{(-6) ^{2}+(1) ^{2} }$

$d=\sqrt{36+1 }$

$d=\sqrt{37}$

Thus length of AD = $\sqrt{37}$