Question

the coordinates of the vertices of triangle ABC are (-1,3),(-3,2) and (5,-1) respectively. Show that the length of the median through the vertex A is
$\sqrt{41 \div 4}$
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Answer 1

Question :

The co-ordinates of the vertices of ∆ ABC are (-1,3),(-3,2)&(5,-1) respectively. Show that the length of the median through the vertex A is √(5) ?

ANSWER

Given : -

The co-ordinates of the vertices of ∆ABC are (-1,3),(-3,2)&(5,-1) respectively.

Required to find : -

• Show that length of the median through the vertex A is √(5)?

Formulae used : -

Distance formula

√{([x]_[2]-[x]_[1])²+([y]_[2]-[y]_[1])²}

Here,

[x]_[1],[x]_[2],[y]_[1]&[y]_[2] are the co-ordinates of the line

----------------------------------

Mid-Point Formula

(x,y) = ({[x]_[1]+[x]_[2]}/{2},{[y]_[1]+[y]_[2]}/{2})

Here,

[x]_[1],[x]_[2],[y]_[1]&[y]_[2] are the co-ordinates of the line whose mid-point is needed to be found.

Solution : -

The co-ordinates of the vertices of ∆ABC are (-1,3),(-3,2)&(5,-1) respectively.

We need to show that length of the median through the vertex A is √(5) .

So,

The diagram looks like the below one !

-: Diagram :-

$\setlength{\unitlength}{30}\begin{picture}(6,6) \linethickness{0.7} \put(1,1){\line(1,0){5}}\put(3.5,1){\line(0,1){4}}\put(1,1){\circle*{0.2}} \put(6,1){\circle*{0.2}}\put(3.5,5){\circle*{0.2}} \put(3.5,1){\circle*{0.2}}\linethickness{0.7}\qbezier(1,1)(1,1)(3.5,5)\qbezier(6,1)(6,1)(3.5,5)\put(3.7,5){ \bf A( - 1,3) }\put(0.5,0.7){ \bf B( - 3,2) }\put(5.5,0.6){ \bf C(5,1) }\put(3,0.6){ \bf D(x,y) }\end{picture}$

From the diagram we can conclude that;

AD is the median of the ∆ABC.

A,B,C are the vertices of ∆ABC.

Here,

Let the co-ordinates be as ;

A = (-1,3) [{x}_{1},{y}_{1} in general]

B = (-3,3) [{x}_{2},{y}_{2} in general]

C = (5,-1) [{x}_{3},{y}_{3} in general]

We need to apply a bit logic here !

If a median is drawn from the vertex A is falls on the line BC.

However, according to properties of a median we know that ;

A median divides the line on which it falls into 2 equal halves.

So,

The end vertex of the median is the mid-point co-ordinates of line BC.

So,

Let's find the mid point co-ordinates of BC.

Using the formula;

(x,y) = ({[x]_[2]+[x]_[3]}/{2},{[y]_[2]+[y]_[3]}/{2})

Substituting the values;

(x,y) = ({-3+5}/{2},{3+1}/{2})

(x,y) = ({2}/{2},{4}/{2})

(x,y) = (1,2)

Hence,

Mid-Point co-ordinates of the line BC is (1,2)

This implies;

The end point co-ordinates of the median BC = (1,2)

Now,

The co-ordinates of the median BC are;

A(-1,3) & D(1,2)

We need to show that length of median AD = √(5).

So,

Using the distance formula;

• √[(x-[x]_[1])²+(y-[y]_[1]²)]

Substituting the values we get;

AD = √[(1-(-1))²+(2-3)²]

AD = √[(1+1)²+(-1)²]

AD = √[(2)²+1]

AD = √[4+1]

AD = √5

Hence,

The length of median through the vertex A is √5

Answer 2

Step-by-step explanation:

AD = √[(1-(-1))²+(2-3)²]

AD = √[(1+1)²+(-1)²]

AD = √[(2)²+1]

AD = √[4+1]

AD = √5.