Tag: division of a line segment

Questions Related to division of a line segment

Find the value of $k$, so that $(2, 1)$ is the midpoint between $(1, k)$ and $(3, 1)$.

  1. $1$

  2. $2$

  3. $3$

  4. $4$


Correct Option: A
Explanation:

We know the midpoint formula between two points.
$\left(\dfrac{x _{1}+x _{2}}{2},\dfrac{y _{1}+y _{2}}{2}\right)$
Substitute the values, we get
$\left(\dfrac{1+3}{2},\dfrac{k+1}{2}\right)$ $=$ $\left(2, 1\right)$
On equating, we get
$\left(\dfrac{k+1}{2}\right)$ $= 1$
$\Rightarrow k + 1 = 2$
$\Rightarrow k = 2 - 1$
$\Rightarrow k = 1$

Find the midpoints between the coordinates $(2, 3)$ and $(1, 0)$

  1. $\left(\dfrac{1}{2},\dfrac{3}{2}\right)$

  2. $\left(\dfrac{3}{2},\dfrac{1}{2}\right)$

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

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


Correct Option: D
Explanation:

Midpoint between the coordinates $(x _1,y _1)$ and $(x _2,y _2)$ is:

$\left( \dfrac { x _{ 1 }+x _{ 2 } }{ 2 } ,\dfrac { y _{ 1 }+y _{ 2 } }{ 2 }  \right)$
Therefore, the midpoint between the coordinates $(2,3)$ and $(1,0)$ is: 
$\left( \dfrac { 2+1 }{ 2 } ,\dfrac { 3+0 }{ 2 }  \right) =\left( \dfrac { 3 }{ 2 } ,\dfrac { 3 }{ 2 }  \right)$
Therefore, midpoint is $\left (\dfrac {3}{2}, \dfrac {3}{2}\right)$.

In the standard $(x,y)$ coordinate plane, what are the coordinates of the midpoint of a line segment whose endpoints are $(-3,0)$ amd $(7,4)$?

  1. $(2,2)$

  2. $(2,3)$

  3. $(5,2)$

  4. $(5,4)$


Correct Option: A
Explanation:

Given two points $(x _{1},y _{1})$ and $(x _{2},y _{2})$, then the coordinates of the midpoint are $\left (\dfrac{x _{1}+x _{2}}{2},\dfrac{y _{1}+y _{2}}{2}\right)$.
In the above case the points are $(-3,0)$ and $(7,4)$. 

Hence, the midpoint is $\left (\dfrac{-3+7}{2},\dfrac{0+4}{2}\right)=(2,2)$.

Points $A(\sqrt {2}, 4), B(6, -\sqrt {3})$ and $C$ are collinear. If $B$ is the midpoint of line segment $AC$, approximately calculate the $(x, y)$ coordinates of point $C$.

  1. $(3.71, 1.13)$

  2. $(3.71, 5.73)$

  3. $(7.41, -7.46)$

  4. $(10.59, -7.46)$


Correct Option: D
Explanation:

Given points $A$ $(\sqrt{2},4)$, $B$ $(6,-\sqrt{3})$ and $C$ are col-linear if $B$ the mid point $AC$. 

 if (A) $(3.71,1.13)$ is point $C$, then point $(x,y) =$ $\dfrac{\sqrt{2}+3.71}{2},\dfrac{4+1.13}{2}$
$\Rightarrow 2.56,2.56$ this not point $B$ 

 if (B) $(3.71,5.73)$ is point $C$, then point $(x,y) =$ $\dfrac{\sqrt{2}+3.71}{2},\dfrac{4+5.73}{2}$
$\Rightarrow 2.56,4.86$ this not point $B$ 

 if (C) $(7.41,-7.46)$ is point $C$, then point $(x,y) =$ $\dfrac{\sqrt{2}+7.41}{2},\dfrac{4-7.46}{2}$
$\Rightarrow 4.41,-1.73$ this not point $B$ 
 if (D) $(10.59,-7.46)$ is point $C$, then point $(x,y) =$ $\dfrac{\sqrt{2}+10.59}{2},\dfrac{4-7.46}{2}$
$\Rightarrow 6,1.73$ this the  point $B (6,1.73)$
 if (E) $(10.59,5.73)$ is point $C$, then point $(x,y) =$ $\dfrac{\sqrt{2}+10.59}{2},\dfrac{4+5.73}{2}$
$\Rightarrow 6,4.86$ this not point $B$. 
So (D) $(10.59,-7.46)$ is mid point $B$ of line $AC$.

A square is formed by the points $(4, 5), (12, 5), (12, -3)$ and $(4, -3)$. Find the coordinates of the point at which the diagonals of the square intersect.

  1. $(8, 5)$

  2. $(9, 6)$

  3. $(8, 1)$

  4. $(12, 1)$


Correct Option: C
Explanation:

We know that the diagonal of square is intersect equal at mid point. 

Then mid point of diagonal $(4,5)$ and $(12,-3)$ is $\left ( \dfrac{12+4}{2} \right ),\left ( \dfrac{5-3}{2} \right )=\left ( \dfrac{16}{2} \right ),\left ( \dfrac{5-3}{2} \right )= (8,1)$
Then mid point of diagonal $(12,5)$ and $(14,-3)$ is $\left ( \dfrac{12+4}{2} \right ),\left ( \dfrac{5-3}{2} \right )=\left ( \dfrac{16}{2} \right ),\left ( \dfrac{5-3}{2} \right )= (8,1)$
Then diagonal intersect at point $(8,1)$.

Given point $A(-3, -8)$, if the midpoint of segment $AB$ is $(1, -5)$, calculate the coordinates of point $B$.

  1. $(5, -2)$

  2. $(4, -2)$

  3. $(-1, -6.5)$

  4. $(-2, -2)$


Correct Option: A
Explanation:

Given, coordinates $A (-3,-8)$ and the mid point of $AB$ is $(1,-5)$

As per Midpoint formula coordinates of mid point $=$ $\dfrac{x _{1}+x _{2}}{2},\dfrac{y _{1}+y _{2}}{2}$
Let the  coordinates of point $B$ be $(x,y)$
Then $(1,-5)=\left [ \left (\dfrac{-3+x}{2}\right),\left (\dfrac{-8+y}{2}\right) \right ]$
Then $\dfrac{-3+x}{2}=1$
$\Rightarrow -3+x=2$
$\Rightarrow x=2+3=5$
Then$\dfrac{-8+y}{2}=-5$
$\Rightarrow -8+y=-10$
$\Rightarrow y=-10+8=-2$
Then coordinates of $B$ is $(5,-2)$.

R is the midpoint of the segment $\bar{PT}$, and $Q$ is the midpoint of line segment $\bar{PR}$. If $S$ is a point between $R$ and $T$ such that the length of segment $\overline{QS}$ is $10$ and the length of segment $\overline{PS}$ is $19$, what is the length of segment $\overline{ST}$?

  1. $13$

  2. $14$

  3. $15$

  4. $16$

  5. $17$


Correct Option: E
Explanation:

Given that $PR=RT$ and $PQ=QR$

Let $QR=PQ=x$ , we get $PR=RT=2x$
Given that $S$ is a point between $R$ and $T$
Given $QS=10$ , $PS=19$
$PS=PQ+QS=PQ+10=19$
$\Rightarrow PQ=9$
Therefore we get $PT=4x=36$
$\Rightarrow ST=PT-PS=36-19=17$

In the $xy$-coordinate plane, the coordinates of three vertices of a rectangle are $\left(1, 5\right)$, $\left(5, 2\right)$ and $\left(5, 5\right)$. What are the coordinates of the fourth vertex of the rectangle?

  1. $\left(1, 2\right)$

  2. $\left(1, 7\right)$

  3. $\left(2, 1\right)$

  4. $\left(2, 5\right)$

  5. $\left(5, 7\right)$


Correct Option: A
Explanation:

The intersection point of diagonals of rectangle is midpoint of each diagonal.

Let the fourth coordinate be $(x,y)$. By the above property we get 
$ \dfrac { x+5 }{ 2 } =\dfrac { 1+5 }{ 2 } $. which implies $x=1$.
$\dfrac { y+5 }{ 2 } =\dfrac { 2+5 }{ 2 } $. which implies $y=2$).
So, the coordinate $(x,y)$ is $(1,2)$.

If $\left (\dfrac {a}{3}, 4\right )$ is the midpoint of the line segment joining $A (-6, 5)$ and $B(-2, 3)$, find $a$.

  1. $-4$

  2. $-12$

  3. $12$

  4. $-6$


Correct Option: B
Explanation:

Given the point A(-6,5) and B(-2,3) $\left( \dfrac { a }{ 3 } ,4 \right) $ is the middle of AB

$\Rightarrow \left( \dfrac { -2-6 }{ 2 } ,\dfrac { 3+5 }{ 2 }  \right) =\left( \dfrac { a }{ 3 } ,4 \right) \ \Rightarrow \dfrac { -8 }{ 2 } =\dfrac { a }{ 3 } \ \therefore a=-\dfrac { 24 }{ 2 } =-12\ $

$A(-3,2)$ and $B(5,4)$ are the end points of a line segment, find the coordinates of the midpoints of the line segment.

  1. $(1,3)$

  2. $(3,3)$

  3. $(1,1)$

  4. $(3,1)$


Correct Option: A
Explanation:

Since $A(-3,2)\equiv(x _1,y _1)$ and $B(5,4)\equiv(x _2,y _2)$ are the end points of a line segment.


Therefore, the coordinates of the midpoints of the line segment is given by:


$(x,y)=\left( \dfrac { { x } _{ 1 }+{ x } _{ 2 } }{ 2 } ,\dfrac { { y } _{ 1 }+{ y } _{ 2 } }{ 2 }  \right)$

$ \\ \Rightarrow (x,y)=\left( \dfrac { -3+5 }{ 2 } ,\dfrac { 2+4 }{ 2 }  \right) \quad $

$\\ \Rightarrow (x,y)=\left( \dfrac { 2 }{ 2 } ,\dfrac { 6 }{ 2 }  \right) =\left( 1,3 \right)$