Question

find the lengths of median of triangle whose vertices are A(-1,1), B(5,-3) and C(3,5)

Answer 1

here's your answer.....

Answer 2

Answer:

Step-by-step explanation:

Using the distance formula, AC=$\sqrt{(3+1)^{2}+(5-1)^{2}}$

AC=$\sqrt{16+16}=\sqrt{32}=4\sqrt{2}$

And BC=$\sqrt{(3-5)^{2}+(5+3)^{2}}$

=$2\sqrt{17}$

Now, BD=DC

Therefore, BD+CD=BC

2CD=BC

CD=$\frac{\sqrt{17}}{2}$

Now, in triangle ACD,

$(AC)^{2}=(AB)^{2}+(CD)^{2}$

$(4\sqrt{2}) ^{2}=(AD)^{2}+(\sqrt{17})^{2}$

$32=(AD)^{2}+17$

$(AD)^{2}=15$

$AD=\sqrt{15}cm$

therefore, the length of the median of the triangle is $\sqrt{15}cm$