Tag: midpoints

Questions Related to midpoints

$A\equiv(0, b), B\equiv(0, 0) $ and $C\equiv(a, 0)$ are the vertices of $\triangle ABC. D, E, F$ are the mid-points of the sides $BC, CA $ and $AB $ respectively. If  $a^{2}+ b^{2} = 20$ then

maths constructions mid-point formula midpoints division of a line segment

  1. $(AD)^{2}=9$

  2. $(BE)^{2}=4$

  3. $(AD)^{2}+(CF)^{2}=25$

  4. $(AD)^{2}+(CF)^{2}=(BE)^{2}$

Show answer
Correct Option: C,D
Explanation:

As $A\equiv\left( 0,b \right) ,B\equiv\left( 0,0 \right) $ and $C\equiv\left( a,0 \right) $ 
Then $D\equiv\left( \cfrac { a }{ 2 } ,0 \right) ,E\equiv\left( \cfrac { a }{ 2 } ,\cfrac { b }{ 2 }  \right) $ and $F\equiv\left( 0,\cfrac { b }{ 2 }  \right) $
Using $a^{ 2 }+b^{ 2 }=20$, we have
${ \left( AD \right)  }^{ 2 }={ \left( \cfrac { a }{ 2 } -0 \right)  }^{ 2 }+{ \left( 0-b \right)  }^{ 2 }=\cfrac { { a }^{ 2 } }{ 4 } +{ b }^{ 2 }\ Similary\,\, {(CF)}^2\ { \left( AD \right)  }^{ 2 }+{ \left( CF \right)  }^{ 2 }={ \left( \cfrac { a }{ 2 } -0 \right)  }^{ 2 }+{ \left( 0-b \right)  }^{ 2 }+{ \left( 0-a \right)  }^{ 2 }+{ \left( \cfrac { b }{ 2 } -0 \right)  }^{ 2 }\ =\cfrac { 5\left( { a }^{ 2 }+{ b }^{ 2 } \right)  }{ 4 } =25={ \left( BE \right)  }^{ 2 }$

If two vertices of a parellelogram are $(3,2)$ and $(-1,0)$ and the diagonals intersect at $(2, -5)$, then the other two vertices are:

maths constructions mid-point formula midpoints division of a line segment

  1. $(1, -10),(5, -12)$

  2. $(1, -12),(5, -10)$

  3. $(2, -10),(5, -12)$

  4. $(1, -10),(2, -12)$

Show answer
Correct Option: B
Explanation:

Let the given vertices be $ A (1,-12)  B (5, -10)  C(a,b) ; D(x,y) $

We know that the diagonals of a parallelogram bisect each other. So,the midpoint of AC $ = (2,-5) $ and of BD $ = (2,-5) $. 

Mid point of two points $ { (x } _{ 1 },{ y } _{
1 }) $ and $ { (x } _{ 2 },{ y } _{ 2 }) $ is  calculated by the formula

$ \left( \frac { { x } _{ 1 }+{ x } _{ 2 } }{ 2 } ,\frac { { y } _{ 1 }+y _{ 2 }

}{ 2 }  \right) $
Hence, midpoint of $ AC = (2,-5) $
$ => \left( \frac {1 + a }{ 2 } ,\frac { -12 + b }{ 2 }  \right) \quad = (2,-5) $
$ => \frac {1 + a }{ 2 } = 2 ; \frac { -12 + b }{ 2 }

= -5 $
$ => 1+ a = 4 ; -12 + b = -10 $
$ a = 3 ; b = 2 $

Hence, C $ =(3,2) $

And, midpoint of $ BD = (2,-5) $
$ => \left( \frac {5 + x}{ 2 } ,\frac { -10 + y }{ 2 }  \right) \quad = (2,-5) $
$ => \frac {5 + x }{ 2 } = 2 ; \frac { -10 + y }{ 2 }

= -5 $
$ => 5 + x = 4 ; -10 + y = -10 $
$ x = -1 ; y = 0 $

$ => D  =(-1,0) $