Geo2.31

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If (2,1),(-5,7),(-5,-5)\, are mid points of sides of a triangle,find the equations of the sides of a triangle.

Let the three vertices are

A(x_{1},y_{1}),B(x_{2},y_{2}),C(x_{3},y_{3})\,

Let the mid point of AB is (2,1)

Hence

x_{1}+x_{2}=4,y_{1}+y_{2}=2\,

Let the mid point of BC is (-5,7)

Hence

x_{3}+x_{2}=-10,y_{3}+y_{2}=14\,

Let the mid point of CA is (-5,-5)

Hence

x_{3}+x_{1}=-10,y_{3}+y_{1}=-10\,

Solving the three six equations we get

A(2,-11),B(2,13),C(-12,1)\,

Now the equations are

y+11={\frac  {13+11}{2-2}}(x-2),y-13={\frac  {1-13}{-12-2}}(x-2),y-1={\frac  {-11-1}{2+12}}(x+12)\,

x-2=0,-14y+182=-12x+24,14y-14=-12x-144\,

x=2,12x-14y+158=0,12x+14y+130=0\,

After simplifying,the final equations are

x=2,6x-7y+79=0,6x+7y+65=0\,


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