Question

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$\bf Question:-$

$\sf {Prove\:using\:co-ordinate\:Geometry\:that,} \\ \sf {All\: three \:Me diums\: of\:any\:triangle\:always\:p as s\:through\:a\:single\;dot (Only\:one\:dot)}$
​

Answer 1

Answer:

Diagram:-

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Solution:-

$\sf In \triangle ABC$

Let,

$\sf A(x_1,y_1)\:,B (x_2,y_2)\:and\:C (x_3,y_3)$

Here ,The three Medians are

$\sf AD,BE,CF$

And,

$\sf D\:is\:the\:centre\:of\:BC$

$\therefore\sf D=\left (\dfrac {x_2+x_3}{2},\dfrac {y_2+y_3}{2} \right)$

$\qquad\qquad\sf {:}\longrightarrow E=\left (\dfrac {x_1+x_3}{2},\dfrac {y_1+y_3}{2}\right)$

$\qquad\qquad\sf {:}\longrightarrow F=\left (\dfrac {x_1+x_2}{2},\dfrac {y_1+y_2}{2}\right)$

Here,

G is the centroid of the triangle

As we know that,

At centroid three medians intersect each other in the ratio of 2:1.

Thus,

Over AD median,

$\qquad\qquad\sf {:}\longrightarrow G=\left (\dfrac {\dfrac {2 (x_2+x_3)}{2}+1 (x_1)}{2+1},\dfrac {\dfrac {2 (y_2+y_3)}{2}+1 (y_1)}{2+1}\right)$

$\qquad\qquad\sf {:}\longrightarrow G=\left(\dfrac {x_2+x_3+x_1}{3},\dfrac {y_2+y_3+y_1}{3}\right)$

$\qquad\qquad\sf {:}\longrightarrow G=\left (\dfrac {x_1+x_2+x_3}{3},\dfrac {y_1+y_2+y _3}{3}\right)$

Over BE median,

$\qquad\qquad\sf {:}\longrightarrow G=\left (\dfrac {\dfrac {2 (x_1+x_3)}{2}+1 (x_2)}{2+1},\dfrac {\dfrac {2 (y_1+y_3)}{2}+1 (y_2)}{2+1}\right)$

$\qquad\qquad\sf {:}\longrightarrow G=\left (\dfrac {\dfrac {2 (x_1+x_3)}{2}+x_2}{2+1},\dfrac {\dfrac {2 (y_1+y_3)}{2}+y_2}{2+1}\right)$

$\qquad\qquad\sf {:}\longrightarrow G=\left (\dfrac {x_1+x_2+x_3}{3},\dfrac {y_1+y_2+y _3}{3}\right)$

Over CF median,

$\qquad\qquad\sf {:}\longrightarrow G=\left (\dfrac {\dfrac {2 (x_1+x_2)}{2}+1 (x_3)}{2+1},\dfrac {\dfrac {2 (y_1+y_2)}{2}+1 (y_3)}{2+1}\right)$

$\qquad\qquad\sf {:}\longrightarrow G=\left (\dfrac {\dfrac {2 (x_1+x_2)}{2}+x_3}{2+1},\dfrac {\dfrac {2 (y_1+y_2)}{2}+y_3}{2+1}\right)$

$\qquad\qquad\sf {:}\longrightarrow G=\left (\dfrac {x_1+x_2+x_3}{3},\dfrac {y_1+y_2+y _3}{3}\right)$

_________________________________

Here,

We see that over All three medians AD,BE,CF

$\qquad\qquad\sf {:}\longrightarrow G=\left (\dfrac {x_1+x_2+x_3}{3},\dfrac {y_1+y_2+y _3}{3}\right)$

Hence

All three medians Namely AD,BE,CF pas s through a Single Dot(G)

Hence Proved