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Questions Related to maths

Find the midpoint of the line segment joining the points  $(1,-1)$ and $(-5,-3)$

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

  1. $(-2,1)$

  2. $(2,1)$

  3. $(-2,-1)$

  4. None of these

Show answer
Correct Option: A
Explanation:
Take $(x _1,y _1)=(1,-1)$ and $(x _2,y _2)=(-5,-3)$.

By midpoint theorem:

$x=\dfrac{x _1+x _2}{2}$ and $y=\dfrac{y _1+y _2}{2}$

Hence, $x=\dfrac{1+(-5)}{2}=-2$ and $y=\dfrac{-1-(-3)}{2}=1$.

So, $(x,y)=(-2,1)$.

The centre of a circle is at $(-6,4)$. If one end of a diameter of the circle is at the origin, then find the other end.

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

  1. $(-12,8)$

  2. $(-12,-8)$

  3. $(12,8)$

  4. None of these

Show answer
Correct Option: A
Explanation:

Take coordinate of other end of diameter as $P(x,y)$.


since, origin divides the diameter of a circle in $2$ equal parts, hence origin will be mid-point of $(0,0)$ and $(x,y)$.


By midpoint theorem:

$x=\dfrac{x _1+x _2}{2}$ and $y=\dfrac{y _1+y _2}{2}$

$=>-6=\dfrac{0+x}{2}=\dfrac{x}{2}$

$=>x=-12$

And,

$=>4=\dfrac{y+0}{2}=\dfrac{y}{2}$

$=>y=8$.

So, $P(x,y)=(-12,8)$

Find the mid point of (3,8) and (9,4).

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

  1. $(5,6)$

  2. $(6,6)$

  3. $(4,4)$

  4. None of the above

Show answer
Correct Option: B
Explanation:

Midpoint formula is given by $\left(\dfrac{x _1+x _2}{2},\dfrac{y _1+y _2}{2} \right)$

So midpoint of $(3,8)$ and $(9,4)$ is $=\left(\dfrac{3+9}{2},\dfrac{8+4}{2} \right)=(6,6)$

Find the mid point of $(4,6)$ and $(2,-6)$.

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

  1. $(3,4)$

  2. $(2,-2)$

  3. $(3,0)$

  4. None of the above

Show answer
Correct Option: C
Explanation:

We know that end points of a line segment is $(a,b)$ and $(c,d)$, then the midpoint of the line segment has the coordinates:

$\dfrac{a+c}{2},\dfrac{b+d}{2}$
Then mid point of line segment $(4,6)$ and $(2,-6)$ is 

$\dfrac{4+2}{2},\dfrac{6-6}{2}$
$\Rightarrow \dfrac{6}{2},\dfrac{0}{2}$
$\Rightarrow (3,0)$

If mid point of the line segment joining (2a, 4) and (-2, 3b) is  (1, 2a + 1), then the values of  a and b are given by

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

  1. $a = 2, b = - 2$

  2. $a = b = 2$

  3. $a= 1 = b$

  4. $a= -2, b = 2$

Show answer
Correct Option: B
Explanation:
Midpoint of any two points $(a.b)$ and $(c,d)$ is given by
$M=\left(\dfrac{a+c}{2},\dfrac{b+d}{2}\right)$
Given points are $(2a, 4)$ and $(-2,3b)$ 
$\therefore M=\left(\dfrac{2a-2}{2},\dfrac{4+3b}{2}\right)$
$\implies (1,2a+1)=\left(a-1,\dfrac{4+3b}{2}\right)$
$\implies a-1=1$ and $\dfrac{4+3b}{2}=2a+1$
$\implies a=2$ and $4a-3b=2$
$\implies a=2, b=2$
Hence, $a=b=2$.

Find the coordinates of the point where the diagonals of the parallelogram formed by joining the points $(-2,-1),(1,0),(4,3)$ and $(1,2)$ meet.

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

  1. $(1,1)$

  2. $(1,3)$

  3. $(5,1)$

  4. None of these

Show answer
Correct Option: A
Explanation:

The vertices of parallelogram in order are $A(-2,-1), B(1,0), C(4,3), D(1,2)$.


So the diagonals will be $AC$ and $BD$.

Since the diagonals of a parallelogram bisects each other, so mid-point of 

$AC$ or $BD$ will be intersection point of diagonals.

Hence by mid-point theorem, mid-point of $AC$ is

$A(-2,-1) \  and \ C(4,3)$

$x=\dfrac{-2+4}{2}=1$ and $y=\dfrac{-1+3}{2}=1$.

so $(1,1)$ is required point.

$P and Q$ are points on the line joining $A(-2,5) and (3,1)$ such that $AP=PQ=QB$ then the mid point of $PQ$ is ?

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

  1. $\dfrac{1}{23}$

  2. $\dfrac{-1}{24}$

  3. $\dfrac{-24}{1}$

  4. $(1,4)$

Show answer
Correct Option: A
Explanation:

Since $AP = PQ = BQ$


hence mid point of PQ will be the mid point of AB.



Thus mid point of AB = [( –2+3)/2 , (5+1)/2 ]





mid point of AB = [1/2 , 3] = mid point of PQ



 

If $P \left( \dfrac{a}{3},\dfrac{b}{2} \right)$ is the mid-point of the line segment joining $A(-4,3)$ and $B(-2,4)$ then $(a,b)$ is 

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

  1. $(-9,7)$

  2. $\left( -3, \dfrac{7}{2} \right)$

  3. $(9,-7)$

  4. $\left( 3, -\dfrac{7}{2} \right)$

Show answer
Correct Option: A
Explanation:
Using the midpoint formula, $\left(x,y\right)=\left(\dfrac{{x} _{1}+{x} _{2}}{2},\dfrac{{y} _{1}+{y} _{2}}{2}\right)$ where ${x} _{1}=-4$,${y} _{1}=3$,${x} _{2}=-2$, ${y} _{2}=4$
Given $P$ is a mid-point of $AB$
$\Rightarrow P\left(\dfrac{a}{3},\dfrac{b}{2}\right)=\left(\dfrac{-4-2}{2},\dfrac{3+4}{2}\right)=\left(-3,\dfrac{7}{2}\right)$
Equating the $x$ and $y$ coordinates, we get
$\Rightarrow \dfrac{a}{3}=-3,\dfrac{b}{2}=\dfrac{7}{2}$
$\Rightarrow a=-9,b=7$
$\therefore \left(a,b\right)=\left(-9,7\right)$

$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) $