Find the length of the median through the ver | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the vertices of the triangle be $A(-1, 3)$,$B(1, -1)$,and $C(5, 1)$.The median through vertex $A$ connects $A$ to the midpoint of the opposite

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) Let the vertices of the triangle be $A(-1, 3)$,$B(1, -1)$,and $C(5, 1)$.The median through vertex $A$ connects $A$ to the midpoint of the opposite side $BC$.Let $M$ be the midpoint of $BC$. The coordinates of $M$ are given by the midpoint formula: $M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{1 + 5}{2}, \frac{-1 + 1}{2} \right) = \left( \frac{6}{2}, \frac{0}{2} \right) = (3, 0)$.The length of the median $AM$ is the distance between $A(-1, 3)$ and $M(3, 0)$.Using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.$AM = \sqrt{(3 - (-1))^2 + (0 - 3)^2} = \sqrt{(4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.Thus,the length of the median is $5$.

Find the length of the median through the vertex $(-1, 3)$ of the triangle having vertices $(-1, 3)$,$(1, -1)$,and $(5, 1)$.

Similar Questions

Prove that $A (1, 7)$,$B (2, 4)$,and $C (5, 5)$ are the vertices of an isosceles right triangle.

The distance of the point $P$ on the $X-$axis from $A(11, 12)$ is $13$ units. The coordinates of $P$ are $\ldots \ldots \ldots \ldots$.

If the point $A (2, -4)$ is equidistant from $P (3, 8)$ and $Q (-10, y)$,find the values of $y$. Also,find the distance $PQ$.

The distance of point $P(2, -3)$ from the $X$-axis is $\ldots \ldots \ldots \ldots .$.

The distance between $A(2,4)$ and $B(-3,4)$ is............

Vedclass Test Series

Exam Paper Generator

Online Exam Module