Question

the midpoint of the sides BC, CA and AB of a triangle ABC are D(2,1) , E (-5,7) and F (-5,-5) respectively. Find te equation of sides of triangle ABC.

Answer 1

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LET,

THE CO-ORDINATES OF POINT A :

=> ( X1 , Y1 )

THE CO-ORDINATES OF POINT B :

=> ( X2 , Y2 )

LET P BE ITS MIDPOINT ,

=> CO - ORDINATES OF POINT P :

=> ( X , Y )

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THEN THE CO-ORDINATES OF POINT P :

=> $x = \frac{x1 + x2}{2}$

=> $y = \frac{y1 + y2}{2}$

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AND THIS IS THE FORMULA FOR MIDPOINT ....

NOW FROM FIG. ( ATTACHMENT )

$f = \frac{x1 + x2}{2}$

$- 5 = \frac{x1 + x2}{2}$

$- 5 \times 2 = x1 + x2$

$- 10 = x1 + x2 \: ..............1$

$f = \frac{y1 + y2}{2}$

$- 5 = \frac{y1 + y2}{2} \\ \\ - 5 \times 2 = y1 + y2 \\ \\ - 10 = y1 + y2 \: .................2\\ \\ d = \frac{x2 + x3}{2} \\ \\ 2 = \frac{x2 + x3}{2} \\ \\ 2 \times 2 = x2 + x3 \\ \\ 4 = x2 + x3.................3\\ \\ d = \frac{y2 + y3}{2} \\ \\ 1 = \frac{y2 + y3}{2} \\ \\ 1 \times 2 = y2 + y3 \\ \\ 2 = y2 + y3 \: .....................4\\ \\ e = \frac{x1 + x3}{2} \\ \\ - 5 = \frac{x1 + x3}{2} \\ \\ - 5 \times 2 = x1 + x3 \\ \\ - 10 = x1 + x3 \: ..................5\\ \\ e = \frac{y1 + y3}{2} \\ \\ 7 = \frac{y1 + y3}{2} \\ \\ 7 \times 2 = y1 + y3 \\ \\ 14 = y1 + y3 \: ................6$
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SUBSTRACT EQUATIONS 1 AND 3 ,

-10 = X1 + X2
-
-10 = X1 + X3
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X2 - X3 = 0

=> X2 = X3

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BUT X2 + X3 = 4

=> X2 + X2 = 4 .................. SINCE X2 = X3

=> 2X2 = 4

=> X2 = 4 / 2

=> X2 = 2 = X3

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PUT X2 = 2 IN EQUATION 1 :

=> -10 = X1 + 2

=> X1 = -10 -2

=> X1 = -12

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SUBSTRACT EQUATION 6 FROM 2

-10 = Y1+Y2
-
+14 = Y1 + Y3
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-24 = Y2 - Y3

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ADD THIS EQUATION WITH EQUATION 4

-24 = Y2 - Y3
+
+2 = Y2 + Y3
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-22 = 2Y2

=> Y2 = -22 / 2

=> Y2 = -11

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PUT Y2 = -11 IN EQUATION 4 :

=> 2 = -11 + Y3

=> 2 + 11 = Y3

=> Y3 = 13

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PUT Y3 = 13 IN EQUATION 6 ;

=> 14 = Y1 + 13

=> Y1 = 14 - 13

=> Y1 = 1

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SO,

CO-ORDINATES OF POINT A :

=> ( X1 , Y1 ) :

=> ( -12 , 1 )

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CO-ORDINATES OF POINT B ;

=> ( X2 , Y2 )

=> ( 2 , -11 )

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CO-ORDINATES OF POINT C :

=> ( X3 , Y3 )

=> ( 2 , 13 )

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TO FIND :

SIDES

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HERE DISTANCE FORMULA SHOULD BE USED TO FIND THE SIDES

DISTANCE FORMULA :

√( X2 - X1 )^2 + ( Y2 - Y1 )^2

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D ( A,B )

√[ 2 - ( -12 ) ]^2 + ( -11 - 1 ) ^ 2

√ ( 2 + 12 ) ^2 + ( -12 ) ^ 2

√ 14^2 + -12^2

√196 + 144

√ 340

2√ 85

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D ( B ,C )

√( 2 - 2 ) ^ 2 + ( -11 - 13 ) ^ 2

√ 0 + -24^2

√ 0 + 576

√576

24

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D ( A , C )

√ [ 2 - ( -12 ) ] ^ 2 + ( 13 - 1 ) ^ 2

√ -14 ^ 2 + 12 ^ 2

√ 196 + 144

√ 340

2 √ 85

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SIDE AB = SIDE AC

SO, IT IS AN ISOSCELES TRIANGLE .....

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THANKS .....

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