Question

If AD is the median of triangle ABC whose vertices are A (-2,6) B(2,1) and C(8,7) find the length of median AD
(I need the answer using the formula of coordinate geometry chapter)

Answer 1

Answer:

Step-by-step explanation:

Since , median of a ∆ meets the side at is mid point

Mid point of BC = ( (x1 + X2 )/2 , (y1 + y2 )/2 ) =( (2+8)/2 , ( 7+1) /2 )= (5,4)

Therefore pt. D = (5,4)

Using distance formula ,

AD = √((X2 - x1 )² + (y2 - y2)² )

= √(-2-5)² + (4-6)² = √ ( 49 + 4 ) = √53 units ( Ans. )

Answer 2

Concept

A line segment connecting a vertex to the midpoint of the opposite side is referred to as a triangle's median.

The mid-point of a line joining the coordinates $\[\left( {{x}_{1,}}{{y}_{1}} \right)\]$ and $\[\left( {{x}_{2,}}{{y}_{2}} \right)\]$ is given as-

$\[=\left( \frac{{{x}_{1}}+{{x}_{2}}}{2},\frac{{{y}_{1}}+{{y}_{2}}}{2} \right)\]$

The length of a line joining the coordinates $\[\left( {{x}_{1,}}{{y}_{1}} \right)\]$ and $\[\left( {{x}_{2,}}{{y}_{2}} \right)\]$ is given as-

$\[=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\]$

Given

The three vertices of a triangle ABC are A(-2,6), B(2,1) and C(8,7).

Find

We have to find the length of the median AD of a triangle ABC.

Solution

Consider the following figure,

The coordinates of mid-point D is given as-

$\[=\left( \frac{2+8}{2},\frac{1+7}{2} \right)\]$

$\[=\left( 5,4 \right)\]$

Now, the length of the median AD is given as-

$\[=\sqrt{{{\left( 5-(-2) \right)}^{2}}+{{\left( 4-6 \right)}^{2}}}\]$

$\[=\sqrt{{{\left( 7 \right)}^{2}}+{{\left( -2 \right)}^{2}}}\]$

$\[=\sqrt{49+4}\]$

$\[=\sqrt{53}\]$

Hence, the length of the median AD is $\[\sqrt{53}\]$ units.

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