Question

a
4.
If A (2, 2), B (-4,-4), C(5,-8) are the vertices of A ABC , then find the length of the median through ‘C'.​

Answer 1

Given :

In ∆ABC,

• A(2, 2)
• B(-4, -4)
• C(5, -8)
• The median is through "C".

To FinD :

The length of the median through "C".

Solution :

Analysis :

Here the concept of mid point formula and distance formula is used. First we have to find the coordinates of the other point of the median. Then by using that point and we are already having the coordinates of vertex C we can find the length of the median using distance formula.

Required Formula :

$\normalsize\boxed{\bf Mid-Point\:Formula=\bigg\lgroup\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\bigg\rgroup}$

$\normalsize\boxed{\bf Distance\:Formula=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$

Explanation :

Let the Median be "CD".

Let the coordinates of D be "(x, y)".

We can find the coordinates of D by finding the midpoint of AB.

• A(2, 2)
• B(-4, -4)

By using the midpoint formula,

$\\ \normalsize\sf (x, y)=\bigg\lgroup\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\bigg\rgroup$

where,

• x₁ = 2
• x₂ = -4
• y₁ = 2
• y₂ = -4

Substituting the values,

$\\ :\implies\normalsize\sf (x, y)=\bigg\lgroup\dfrac{2+(-4)}{2},\dfrac{2+(-4)}{2}\bigg\rgroup$

$\\ :\implies\normalsize\sf (x, y)=\bigg\lgroup\dfrac{2-4}{2},\dfrac{2-4}{2}\bigg\rgroup$

$\\ :\implies\normalsize\sf (x, y) =\bigg\lgroup\dfrac{-2}{2},\dfrac{-2}{2}\bigg\rgroup$

$\\ :\implies\normalsize\sf (x, y) =\bigg\lgroup\cancel{\dfrac{-2}{2}},\cancel{\dfrac{-2}{2}}\bigg\rgroup$

$\\ \normalsize\therefore\boxed{\bf (x, y)=(-1,-1)}$

Coordinates of D(-1, -1).

Now the Distance between C & D :

• C(5, -8)
• D(-1, -1)

Using distance formula,

$\\ \normalsize\sf Distance\:Formula=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

where,

• x₁ = 5
• x₂ = -1
• y₁ = -8
• y₂ = -1

Substituting the values,

$\\ :\implies\normalsize\sf Distance\:Formula=\sqrt{(-1-5)^2+(-1-(-8))^2}$

$\\ :\implies\normalsize\sf Distance\:Formula=\sqrt{(-1-5)^2+(-1+8)^2}$

$\\ :\implies\normalsize\sf Distance\:Formula=\sqrt{(-6)^2+(7)^2}$

$\\ :\implies\normalsize\sf Distance\:Formula=\sqrt{36+49}$

$\\ :\implies\normalsize\sf Distance\:Formula=\sqrt{85}$

$\\ \normalsize\therefore\boxed{\bf Distance\:Formula=\sqrt{85}\ units.}$

The length of the median through C is √85 units.

(for more reference refer to the attachment)

Answer 2

Given Question :-

• If A (2, 2), B (-4,-4), C(5,-8) are the vertices of triangle ABC , then find the length of the median through ‘C'.

$\begin{gathered} \large{\bf{\green{\underline{GiVeN\::}}}} \end{gathered}</p><p>$

• A (2, 2), B (-4,-4), C(5,-8) are the vertices of A ABC.

$\begin{gathered} \Large{\bf{\pink{\underline{To\:FiNd\::}}}} \end{gathered}$

• The length of the median through ‘C'.

Formula Used :-

☆ Midpoint Formula :-

Let us consider a line segment joining the points

$\sf \: A(x_1,y_1) \: and \: B(x_2,y_2)$

and Let C (x, y) is the midpoint of AB, then coordinates of midpoint C is given by

$\bf \:( x, y) = (\dfrac{x_1+x_2}{2} , \dfrac{y_1+y_2}{2} )$

☆Distance Formula :-

Let us consider a line segment joining the points

$\sf \: A(x_1,y_1) \: and \: B(x_2,y_2)$

then distance AB is given by

$\bf\implies \:AB = \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} }$

$\begin{gathered} \Large{\bf{\purple{\underline{CaLcULaTIoN\::}}}}\end{gathered}$

A (2, 2), B (-4,-4), C(5,-8) are the vertices of triangle ABC.

☆Let CD be the median through C intersecting AB at D.

So, D is the midpoint of AB.

☆Let coordinates of D be (x, y).

Then, Coordinates of D using midpoint Formula is

$\bf \:( x, y) = (\dfrac{2 - 4}{2} , \dfrac{2 - 4}{2} ) = ( - 1, - 1)$

So coordinates of D are (- 1, - 1).

Now, we have to find the length of CD.

Coordinates of C = (5, - 8)

Coordinates of D = (- 1, -1)

Length of median AD using distance formula is

$\sf \: Length \: AD = \sqrt{ {(5 + 1)}^{2} + {( - 8 + 1)}^{2} }$

$\sf \: ⟼AD = \sqrt{ {(6)}^{2} + {( - 7)}^{2} }$

$\sf \: ⟼AD = \sqrt{36 + 49}$

$\sf \: ⟼AD = \sqrt{85} \: units$

${ \boxed {\bf{Hence, length \: of \: median = \sqrt{85} \: units}}}$