Question

The coordinates of Vertices A , B , C of a triangle ABC are (0 , -2) , (4 , 1) and (0 , 4) respectively. Find the length of median through B.
(a) 4
(b) 5
(c) 6
(d) None of these ​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

• The coordinates of vertices A , B , C of a triangle ABC are (0 , -2) , (4 , 1) and (0 , 4) respectively.

We know,

Median is a line segment drawn from the vertex which bisects the opposite side of the triangle.

As median is drawn from vertex B of triangle ABC, so let assume that median yhrough B bisects the side AC at D

So, BD be the median and D is the midpoint of AC.

Let assume that coordinates of D be (a, b).

We know

Mid-point formula

Let P(x₁, y₁) and Q(x₂, y₂) be two points in the coordinate plane and R(x, y) be the mid-point of PQ. Then, the coordinates of R will be:

$\sf\implies R = \bigg(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\bigg)$

So, on substituting the values, we get

$\rm :\longmapsto\:(a,b) = \bigg(\dfrac{0 + 0}{2}, \dfrac{ - 2 + 4}{2} \bigg)$

$\rm :\longmapsto\:(a,b) = \bigg(0, \dfrac{2}{2} \bigg)$

$\rm :\longmapsto\:(a,b) = (0,1)$

So, Coordinates of D be (0, 1).

Now, We know

Distance Formula

Let P(x₁, y₁) and Q(x₂, y₂) be two points in the coordinate plane, then the distance between P and Q is

$\rm\implies \:\boxed{\tt{ \sf \: PQ= \sqrt{ {(x_{1} - x_{2}) }^{2} + {(y_{2} - y_{1})}^{2} }}}$

So, Length of median BD having coordinates B(4, 1) and D(0, 1) is

$\rm :\longmapsto\:BD = \sqrt{ {(4 - 0)}^{2} + {(1 - 1)}^{2} }$

$\rm :\longmapsto\:BD = \sqrt{ {4}^{2} + {0}^{2} }$

$\rm :\longmapsto\:BD = \sqrt{ {4}^{2}}$

$\bf\implies \:BD \: = \: 4 \: units$

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1. Section formula.

Let P(x₁, y₁) and Q(x₂, y₂) be two points in the coordinate plane and R(x, y) be the point which divides PQ internally in the ratio m₁ : m₂. Then, the coordinates of R will be:

$\sf\implies R = \bigg(\dfrac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}, \dfrac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}\bigg)$

2. Centroid of a triangle

Centroid of a triangle is the point where the medians of the triangle meet.

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of a triangle. Let R(x, y) be the centroid of the triangle. Then, the coordinates of R will be:

$\sf\implies R = \bigg(\dfrac{x_{1}+x_{2}+x_{3}}{3}, \dfrac{y_{1}+y_{2}+y_{3}}{3}\bigg)$

Answer 2

If the coordinates of Vertices of a triangle ABC are (0 , -2) , (4 , 1) and (0 , 4) respectively then length of median through B is 4 units

Given Data is :

Vertices of Triangle ABC are A(0 , -2) , B( 4 ,1 ) and C (0 , 4)

"Median from a vertex bisect the opposite side"

Hence BD Median through B will bisect AC

D is the mid point of AC

Calculate the mid point of AC

D = (0 + 0)/2 , (-2 + 4)/2

=> D = 0 , 1

Length of Median BD   to be calculated using Distance formula

B = (4 ,1 ) ,  D = ( 0 , 1)

Length of BD = $\sqrt{(4-0)^2+(1-1)^2} = 4$

Hence,   the length of median through B is 4 units

Correct option is a) 4