Question

The coordinate of the vertices of triangle ABC are A(7,2) ,B(9,10) and C (1,4) if E and F are mid points of AB and AC respectively prove that EF=1/2 BC

Answer 1

hope this is correct

Answer 2

To prove $EF=1/2 BC$.

Step-by-step explanation:

Given:

The coordinate of the vertices of the triangle $ABC$ are

$A(7,2)$

$B(9,10)$

$C (1,4)$

Midpoint of $AB$ and $AC$  $=E,F$

To Find:

To prove that $EF=1/2 BC$.

Formula used:

$E(x)=\frac{x_{1}+y_{1} }{2} +\frac{x_{2}+y_{2} }{2}$, $F(x)=\frac{x_{1}+y_{1} }{2} +\frac{x_{2}+y_{2} }{2}$

Solution:

Step 1:

Find the coordinate points $E$.

Now apply$E(x)=\frac{x_{1}+y_{1} }{2} +\frac{x_{2}+y_{2} }{2}$

$A(7,2) ,B(9,10)$

$(x_{1} y_{1} )=\frac{9+7}{2}+\frac{10+2}{2}$

$(x_{1} y_{1} )=(8,6)$

Step 2:

Now find the coordinate points $F$

Now apply $F(x)=\frac{x_{1}+y_{1} }{2} +\frac{x_{2}+y_{2} }{2}$

$C (1,4), A(7,2)$

$(x_{2} ,y_{2} )=\frac{7+1}{2},\frac{2+4}{2}$

$(x_{2} ,y_{2} )=(4,3)$

Step 3:

Find the length $EF$.

$EF=\sqrt{(x_{1}+y_{1} )^{2}+(x_{2}+y_{2})^{2}$

$EF=\sqrt{(8-4 )^{2}+(6-3)^{2}$

$EF=\sqrt{(4 )^{2}+(3)^{2}$

$EF=\sqrt{16+9}$

$EF=\sqrt{25}$

$EF=5unit$ ⇒$(1)$

Step 4:

Now find the length $BC$

$BC=\sqrt{(x_{1}+y_{1} )^{2}+(x_{2}+y_{2})^{2}$

$BC=\sqrt{(9-1 )^{2}+(10-4)^{2}$

$BC=\sqrt{(8)^{2}+(6)^{2}$

$BC=\sqrt{64+36}$

$BC=\sqrt{100}$

$BC=10unit$⇒$(2)$

From the equation (1) and (2), we get

$EF=1/2 BC$

Hence proved.

#SPJ2