If $A (-1, 3), B (1, -1)$ and $C (5, 1)$ are the vertices of a triangle ABC, find the length of the median through A.
Given:
$A (-1, 3), B (1, -1)$ and $C (5, 1)$ are the vertices of a triangle ABC.
To do:
We have to find the length of the median through A.
Solution:
Let $D$ be the mid-point of BC.
This implies, using mid-point formula,
Coordinates of $D=(\frac{5+1}{2}, \frac{1-1}{2})=(3 ,0)$
We know that,
The distance between two points \( \mathrm{A}\left(x_{1}, y_{1}\right) \) and \( \mathrm{B}\left(x_{2}, y_{2}\right) \) is \( \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \).
Therefore,
The length of the median \( \mathrm{AD}=\sqrt{(3+1)^{2}+(0-3)^{2}}=\sqrt{(4)^{2}+(-3)^{2}} \)
\( =\sqrt{16+9} \)
\( =\sqrt{25} \)
\( =5 \) units
The length of the median through A is $5$ units.
Related Articles Find the lengths of the medians of a triangle whose vertices are $A (-1, 3), B (1, -1)$ and $C (5, 1)$.
Find the lengths of the medians of a $\triangle ABC$ having vertices $A (0, -1), B (2, 1)$ and $C (0, 3)$.
Find the area of a triangle whose vertices are$(1, -1), (-4, 6)$ and $(-3, -5)$
The vertices of $\triangle ABC$ are $(-2, 1), (5, 4)$ and $(2, -3)$ respectively. Find the area of the triangle and the length of the altitude through $A$.
Find the centroid of the triangle whose vertices are:$(1, 4), (-1, -1), (3, -2)$
$A (3, 2)$ and $B (-2, 1)$ are two vertices of a triangle ABC whose centroid $G$ has the coordinates $(\frac{5}{3}, −\frac{1}{3})$. Find the coordinates of the third vertex $C$ of the triangle.
If a vertex of a triangle be $(1, 1)$ and the middle points of the sides through it be $(-2, 3)$ and $(5, 2)$, find the other vertices.
If the coordinates of the mid-points of the sides of a triangle are $(1, 1), (2, -3)$ and $(3, 4)$, find the vertices of the triangle.
If the points $A (a, -11), B (5, b), C (2, 15)$ and $D (1, 1)$ are the vertices of a parallelogram $ABCD$, find the values of $a$ and $b$.
$A (4, 2), B (6, 5)$ and $C (1, 4)$ are the vertices of $\triangle ABC$.The median from A meets BC in D. Find the coordinates of the point D.
$A (4, 2), B (6, 5)$ and $C (1, 4)$ are the vertices of $\triangle ABC$.Find the coordinates of point P on AD such that $AP : PD = 2 : 1$.
Find the area of the triangle whose vertices are:(i) $(2, 3), (-1, 0), (2, -4)$(ii) $(-5, -1), (3, -5), (5, 2)$
If the points $A( -2,\ 1) ,\ B( a,\ b)$ and $C( 4,\ -1)$ are collinear and $a-b=1$, find the values of $a$ and $b$.
Show that the points $A (5, 6), B (1, 5), C (2, 1)$ and $D (6, 2)$ are the vertices of a square.
Find the area of quadrilateral ABCD, whose vertices are: $A( -3,\ -1) ,\ B( -2,\ -4) ,\ C( 4,\ -1)$ and$\ D( 3,\ 4) .$
Kickstart Your Career
Get certified by completing the course
Get Started