Tag: division of a line segment

Questions Related to division of a line segment

The point on X-axis which is equidistant from the point (3, 5) and (4, 2)

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

  1. (-6, 0)

  2. (-7, 0)

  3. (7, 0)

  4. None of these

Show answer
Correct Option: B
Explanation:

Let the point on x-axis equidistant from (3,5) and (4,2) be (x,0),


Then, Distance of point from (3,5) = DIstance of point from (4,2)
$\rightarrow \sqrt{(3-x)^2+(5-0)^2} = \sqrt{(4-x)^2+(2-0)^2}$
$\Rightarrow 9+x^2-6x+25=16+x^2-8x+4$
$\rightarrow 2x = -14$
$\rightarrow x = -7$
$\rightarrow $ Point is $(-7,0)$

Thus, B is the correct answer.

If $A(a, b)$ and $B(0, 0)$ are two fixed points. $M _1$ is the mid point of $\overline{AB}$, $M _2$ is the mid point of $\overline{AM _1}$, $M _3$ is the mid point of $\overline{AM _2}$ and so on, then $M _5$ is?

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

  1. $\left(\dfrac{7a}{8}, \dfrac{7b}{8}\right)$

  2. $\left(\dfrac{15a}{16}, \dfrac{15b}{16}\right)$

  3. $\left(\dfrac{31a}{32}, \dfrac{31b}{32}\right)$

  4. $\left(\dfrac{63a}{64}, \dfrac{63b}{64}\right)$

Show answer
Correct Option: A

The co-ordinates of the mid point of segment $KR$, where $K(2.5, -4.3)$ and $R(-1.5, 2.7)$, are

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

  1. $(0.5, 0.8)$

  2. $(-0.5, -0.8)$

  3. $(-0.5, 0.8)$

  4. $(0.5, -0.8)$

Show answer
Correct Option: D
Explanation:

Given that $KR$ is a line segment joining the points $K=(2.5, -4.3)$ and $R=(-1.5, 2.7)$ 

We know that the co-ordinate of the midpoint $M$ of a line joining the points $(x _1, y _1)$ and $(x _2,y _2)$ is given by 

$M=\left( \dfrac { { x } _{ 1 }+{ x } _{ 2 } }{ 2 }, \dfrac { { y } _{ 1 }+{ y } _{ 2 } }{ 2 }  \right)$
Let $M _{KR}$ be the midpoint of the line segment $KR$

 ${ \therefore \quad M } _{ KR }=\left( \dfrac {2.5-1.5}{2}, \dfrac{-4.3+2.7}{2}  \right) =(0.5,-0.8)$

Hence, option D is correct.

Two points $(a, 3)$ and $(5, b)$ are the opposite vertices of a rectangle. If the coordinates $(x, y)$ of the other two vertices satisfy the relation $y = 2x + c$ where $c^{2}+ 2a -b =0$ then the value $c$ can be

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

  1. $2\sqrt{2}+1$

  2. $2\sqrt{2}-1$

  3. $1-2\sqrt{2}$

  4. $-1-2\sqrt{2}$

Show answer
Correct Option: A,C
Explanation:

Mid-point of the other vertices is also the mid-point of the given vertices and hence satisfies the given relation
So, $\cfrac { b+3 }{ 2 } =2\left( \cfrac { a+5 }{ 2 }  \right) +c$
$\Rightarrow 2a+2c-b+7=0$
Also, ${ c }^{ 2 }+2a-b=0$
$\Rightarrow { c }^{ 2 }-2c-7=0\Rightarrow c=1\pm 2\sqrt { 2 } $

The coordinates of the centre of a circle are $(-6,1.5)$. If the ends of a diameter are $(-3,y)$ and $(x, -2)$ then:

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

  1. $x= 9, y=5$

  2. $x=5, y= -9$

  3. $x=-9, y=5$

  4. $x=-9, y=-5$

Show answer
Correct Option: C
Explanation:

The centre of the circle lies at the mid point of the diameter.
Mid-point of two points $ { (x } _{ 1 },{ y } _{ 1 }) $ and $ { (x } _{ 2 },{ y } _{
2 }) $ is  calculated by the formula $ \left( \dfrac { { x } _{ 1 }+{ x
} _{ 2 } }{ 2 } ,\dfrac { { y } _{ 1 }+y _{ 2 } }{ 2 }  \right) $
Using this formula, mid point of $ (3,y) $ and $ (x,2) = \left( \dfrac { -3 + x }{ 2 } ,\dfrac { y - 2 }{ 2 } \right) $
So,$\left( \dfrac { -3 + x }{ 2 } ,\dfrac { y - 2 }{ 2 }  \right) = (-6, 1.5) $
$ \Rightarrow  - 3 + x = -6 \times 2 $ and $ y - 2 = 1.5 \times 2 $
$\Rightarrow x = - 9 ; y = 5 $

Mid-point of the line-segment joining the points $(-5,4)$ and $(9, -8)$ is:

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

  1. $(-7,6)$

  2. $(2, -2)$

  3. $(7,-6)$

  4. $(-2, -2)$

Show answer
Correct Option: B
Explanation:

Midpoint of two points $ { (x } _{ 1 },{ y } _{ 1 }) $ and $ { (x } _{ 2 },{ y } _{
2 }) $ is calculated by the formula $ \left( \cfrac { { x } _{ 1 }+{ x} _{ 2 } }{ 2 } ,\cfrac { { y } _{ 1 }+y _{ 2 } }{ 2 }  \right) $
Using this formula, mid - point of the line-segment joining the points $(5,4)$ and $(9,8)$ is: $= \left( \cfrac { -5 + 9 }{ 2 } ,\cfrac { 4- 8 }{ 2 }  \right)  = (2,-2) $

Three consecutive vertices of a parallelogram are $(1, -2)$, $(3,6)$ and $(5,10)$. The coordinates of the fourth vertex are:

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

  1. $(-3,2)$

  2. $(2, -3)$

  3. $(3,2)$

  4. $(-2, -3)$

Show answer
Correct Option: C
Explanation:

Let the fourth vertex D $ = (x,y) $
We know that the diagonals of a parallelogram bisect each other. So,the

midpoint of AC is same as the mid point of BD.

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) $

So, midpoint of $ AC = $ Mid point of $ BD $

$ => \left( \frac { 1+5 }{ 2 } ,\frac { -2+10 }{ 2 }  \right) \quad =

\left( \frac { 3+x }{ 2 } ,\frac { 6 +y }{ 2 }  \right) \quad $

$ => \left( \frac { 6 }{ 2 } ,\frac { 8 }{ 2 }   \right) \quad =

\left( \frac { 3+x }{ 2 } ,\frac { 6 +y }{ 2 }  \right) \quad $

$ => 3+x=6 ; 6 + y = 8 $

$ => x = 3 ; y = 2 $

Hence, $ D = (3,2) $

Find the coordinates of the centre of a circle, if the coordinates of the end points of a diameter being $(-3,8)$ and $(5,6)$.

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

  1. $(-1,7)$

  2. $(2,7)$

  3. $(-2,7)$

  4. $(1,7)$

Show answer
Correct Option: D
Explanation:

The centre of the circle lies at the mid point of the diameter.

Mid point of two points $ { (x } _{ 1 },{ y } _{ 1 }) $ and $ { (x } _{ 2 },{ y } _{

2 }) $ is  calculated by the formula $ \left( \dfrac { { x } _{ 1 }+{ x

} _{ 2 } }{ 2 } ,\dfrac { { y } _{ 1 }+y _{ 2 } }{ 2 }  \right) $
Using this formula, mid point of the diameter $= \left( \dfrac { -3 + 5}{ 2 } ,\dfrac { 8 + 6 }{ 2 }  \right) = (1,7) $

The mid-point of a line segment is $(5,8)$. If one end point is $(3,5)$, find the second end point

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

  1. $(-7,11)$

  2. $(7,-11)$

  3. $(7,11)$

  4. $(-7,-11)$

Show answer
Correct Option: A,C
Explanation:

Mid point of two points $ { (x } _{ 1 },{ y } _{ 1 }) $ and $ { (x } _{ 2 },{ y } _{

2 }) $ is calculated by the formula $ \left( \dfrac { { x } _{ 1 }+{ x

} _{ 2 } }{ 2 } ,\dfrac { { y } _{ 1 }+y _{ 2 } }{ 2 }  \right) $

Let the other point be $ (x,y) $
So, $=\left( \dfrac {3 + x }{ 2 } ,\dfrac { 5 + y }{ 2 }  \right) \quad =\quad (5,8)

$
$ => \frac {3 + x }{ 2 } = 5 $ and  $ \dfrac { 5 + y } {2} = 8 $
$ => x = 7 , y = 11 $
So, the second end point is $ (7,11) $

The vertices of a parallelogram are $(3, -2)$, $(4,0)$, $(6, -3)$ and $(5, -5)$. The diagonals intersect at the point M. The coordinates of the point M are:

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

  1. $\begin{pmatrix} \frac { 9 }{ 2 },-\frac { 5 }{ 2 } \end{pmatrix}$

  2. $\begin{pmatrix} \frac { 7 }{ 2 },-\frac { 5 }{ 2 }\end{pmatrix}$

  3. $\begin{pmatrix} \frac { 7 }{ 2 },-\frac { 3 }{ 2 }\end{pmatrix}$

  4. None of these

Show answer
Correct Option: A
Explanation:

The diagonals of a parallelogram bisect each other. Hence M is the mid point of the vertices $ (3,-2) ; (6,-3) $ or of the vertices  $ (4,0) ; (5,-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) $
Using this formula, mid point of $ (3, -2) , (6, -3) = \left( \frac { 3\quad +6 }{ 2 } ,\frac { -2-3 }{ 2 }  \right) =\left( \frac { 9 }{ 2 } ,\frac { -5 }{ 2 }  \right) $