Question

FInd the length of median AD and BE of triangle ABC whose vertices are A(1,-1), B(0,4) and C(-5,3).

Answer 1

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Answer 2

Answer:

$\text{Length of median AD is }\sqrt{\frac{130}{4}}units$

$\text{Length of median BE is }\sqrt{13}units$

Step-by-step explanation:

Given the vertices of triangle ABC A(1,-1), B(0,4) and C(-5,3).

we have to find the length of median AD and BE.

As, AD and BE are the median of triangle ABC

⇒ D and E are the mid-point of BC and AC respectively.

From mid-point formula

$\text{The coordinates of mid-point of line segment joining the points }(x_1,y_1) \text{ and } (x_2,y_2) are$

$(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

$\text{The coordinates of D is }(\frac{0-5}{2},\frac{4+3}{2})=(\frac{-5}{2},\frac{7}{2})$

$\text{The coordinates of E is }(\frac{1-5}{2},\frac{-1+3}{2})=(-2,1)$

By distance formula

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$\text{Length of median AD is }$

$AD=\sqrt{(\frac{-5}{2}-1)^2)^2+(\frac{7}{2}+1)^2}$

$=\sqrt{(\frac{-7}{2})^2+(\frac{9}{2})^2}=\sqrt{\frac{130}{4}}units$

$\text{Length of median BE is }$

$BE=\sqrt{(0+2)^2+(4-1)^2}$

$=\sqrt{4+9}=\sqrt{13}units$