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Midpoints - class-XI

Attempted 0/102 Correct 0 Score 0

If $O(0,4)$ and $P(0,-4)$, are the co-ordinates of the line segment $OP$ then co-ordinate of its midpoint are

  1. $(0,-4)$

  2. $(0,4)$

  3. $(-4,0)$

  4. $(0,0)$


Correct Option: D
Explanation:

Midpoint of a line segment having coordiantes $\left({x} _{1},{y} _{1}\right)$ and $\left({x} _{2},{y} _{2}\right)$ is $\left(\dfrac{{x} _{1}+{x} _{2}}{2},\dfrac{{y} _{1}+{y} _{2}}{2}\right)$

$\therefore $ Modpoint of $OP=\left(\dfrac{0+0}{2},\dfrac{4+-4}{2}\right)$
$=\left(0,0\right)$

Find the mid point of $(9,5)$ and $(3,7)$

  1. $(6,6)$

  2. $(12,12)$

  3. $(2,2)$

  4. $(1,1)$


Correct Option: A
Explanation:

Given points $(9,5),(3,7)$

Mid point is given as $\left(\dfrac{x _1+x _2}2,\dfrac{y _1,y _2}{2}\right)\\left(\dfrac{9+3}{2},\dfrac{5+7}{2}\right)\\left(\dfrac{12}{2},\dfrac{12}{2}\right)=(6,6)$

The mid point of $(8,3)$ and $(4,9)$ is 

  1. $(6,6)$

  2. $(4,4)$

  3. $(9,9)$

  4. $(2,2)$


Correct Option: A
Explanation:

The given points are $(8,3)$ and $(4,9)$

The mid point is given as$ \left(\dfrac {8+4}2,\dfrac {3+9}2\right)\(6,6)$

The mid point of $(-1,-3)$ and $(3,7)$

  1. (1, 2) 

  2. (0, 2) 

  3. (0, 4)

  4. (2 ,2)


Correct Option: A
Explanation:

Given points $(-1,-3),(3,7)$

Mid point is given as $\left(\dfrac {-1+3}{2},\dfrac {-3+7}{2}\right)=(1,2)$

The mid point of $(4,9)  $ and $(8,3)$ is 

  1. $(6,6)$

  2. $(5,7)$

  3. $(-6,-6)$

  4. None.


Correct Option: A
Explanation:

The mid point of $(4,9)  $ and $(8,3)$ is given as 

$\left(\dfrac{4+8}2,\dfrac{9+3}2\right)\\left(\dfrac {12}2,\dfrac{12}2\right)=(6,6)$

The mid point of $(2,3)$ and $(8,9)$ is 

  1. $(5,6)$

  2. $(2,8)$

  3. $(5,7)$

  4. $(4,6)$


Correct Option: A
Explanation:

The points are $(x _1,y _1)=(2,3)$ and $(x _2,y _2)=(8,9)$


The mid point is given as $\left(\dfrac{2+8}2,\dfrac {3+9}2\right)$
                                
$\left(\dfrac {10}2,\dfrac {12}2\right)=(5,6)$

The mid point of $(3,4)$ and $(1,-2)$

  1. (2,1)

  2. (1,2)

  3. (2,-1)

  4. (1,-2)


Correct Option: A
Explanation:

The points are $(3,4)$ and $(1,-2)$


The mid point of $(3,4)$ and $(1,-2)$ is given by 

$\left(\dfrac {x _1+x _2}2,\dfrac {y _1+y _2}2\right)\\\left(\dfrac{3+1}2,\dfrac {4-2}2\right)=(2,1)$

The mid-point of the line segment joining $( 2a, 4)$ and $(-2, 2b)$ is $(1, 2a + 1 )$. The values of $a$ and $b$ are 

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

  2. $a = 2, b = -3$

  3. $a = 3, b = - 2$

  4. $a =2, b = 3$


Correct Option: D
Explanation:

Midpoint of two points $ =\left( \cfrac { { x } _{ 1 }+{ x} _{ 2 } }{ 2 } ,\cfrac { { y } _{ 1 }+y _{ 2 } }{ 2 }  \right) $
Given, midpoint of $ (2a,4) $ and $ (-2,2b) = (1,2a+1) $
$ => \left(\cfrac { 2a-2 }{ 2 } ,\cfrac { 4+2b }{ 2 }\right)= (1,2a+1) $
$ => \cfrac { 2a-2 }{ 2 } = 1 ; \cfrac { 4+2b }{ 2 } = 2a + 1 $
$ => 2a -2 = 2 $
$=> a = 2 $

And, $ \cfrac { 4+2b }{ 2 } = 2a + 1 $
$=> \cfrac { 4+2b }{ 2 } = 2(2) + 1 = 5 $
$ => 4 + 2b = 10 $
$ => 2b = 6 $
$=> b = 3 $

The point which lies in the perpendicular bisector of the line segment joining the points A (-2, -5)  and B (2,5) is 

  1. (0, 0)

  2. (0, 2)

  3. (2, 0)

  4. (-2, 0)


Correct Option: A
Explanation:

A perpendicular bisector of a line segment, passed through its midpoint.

If C is the point on AB, through which its perpendicular bisector passes, then C $ = $ mid point of AB.

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 AB $= \left( \frac { -2+2 }{ 2 } ,\frac { -5+5 }{ 2 }  \right)

\quad =\quad (0,0) $



If Q$\displaystyle \left ( \frac{a}{3},4 \right )$ is the mid-point of the line segment joining the points A(-6,5) and B(-2,3), then the value of 'a' is

  1. 4

  2. -6

  3. -8

  4. -12


Correct Option: D
Explanation:

The co-ordinates of the mid-point  of the line segment joining the point $P(x _1,y _1),Q(x _2,y _2)$$=\left(\dfrac{x _1+x _2}{2},\dfrac{y _1+y _2}{2}\right)$

$\Rightarrow \left(\dfrac{a}{3},4\right)=\left(\dfrac{-6+(-2)}{2},\dfrac{5+3}{2}\right)$
$\Rightarrow \left(\dfrac{a}{3},4\right)=\left(\dfrac{-8}{2},4\right)$
X co-ordinates
$\Rightarrow \dfrac{a}{3}=\dfrac{-8}{2}$
$\Rightarrow a=\dfrac{-8\times 3}{2}=-12$

A triangle has vertices A(1,-1) B(2,4) and C(6,0) The length of the median from A is

  1. 3

  2. $\displaystyle 3\sqrt{2}$

  3. $\displaystyle 2\sqrt{3}$

  4. $\displaystyle 2\sqrt{2}$


Correct Option: B
Explanation:

Midpoint of BC = L = (4, 2)
$\displaystyle \therefore AL=\sqrt{\left ( 1-4 \right )^{2}+\left ( -1-2 \right )^{2}}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$

The midpoint of the line segment between P$\displaystyle _{1}$ (x, y) and P$\displaystyle _{2}$ (-2, 4) is P$\displaystyle _{m}$ (2, -1). Find the coordinate.

  1. (6, -5)

  2. (5, -6)

  3. (6, -6)

  4. (-6, 6)


Correct Option: C
Explanation:

Given,

$P _m(2,-1), P _1(x,y)$ and $P _2(-2,4)$

$(2,-1)=\left(\dfrac{x-2}{2}, \dfrac{y+4}{2}\right)$   ..... By midpoint formula 

$\therefore 2=\dfrac{x-2}{2}$
$=>4=x-2$
$=>x=6$

And,
$-1=\dfrac{y+4}{2}$
$=>-2=y+4$
$=>y=-6$

$\therefore (x,y)=(6,-6)$

If (-2, -4) is the midpoint of (6, -7) and (x, y) then the values of x and y are

  1. x = 2, y = 1

  2. x = -10, y = -1

  3. x = 10, y = -1

  4. x = -8 , y = -1

  5. none of these


Correct Option: C
Explanation:

Since, $(-2, -4)$ is the midpoint of $(6, -7)$ and $(x,y).$
$\Rightarrow \dfrac{x+6}2=-2\Rightarrow x=-10$
and $ \dfrac{y-7}2=-4\Rightarrow y=-1$
Option D is correct.

In the xy-plane, find the mid point of the line segment joining the points $\left( 5,9 \right) $ and $\left( 7,11 \right) $.

  1. $(1.5, 2)$

  2. $(6, 10)$

  3. $(5.5, 5)$

  4. $(6, -3.5)$


Correct Option: B
Explanation:

Given line segment point $(5,9)$ and $(7,11)$, then mid point as per section formula: 

$(x,y)=$ $\left ( \dfrac{7+5}{2} ,\dfrac{11+9}{2} \right )$
$=$ $\left ( \dfrac{12}{2} ,\dfrac{20}{2}\right)$
$=$ $ (6,10)$

The coordinates of points $P(-2, 2), Q(3, 2) $ and $R(3, -2)$ are the vertices of a rectangle $PQRS$`. What are the coordinates of S? 

  1. $(-3., -2)$

  2. $(-2, - 2)$

  3. $(3, 2)$

  4. $(2, 2)$


Correct Option: B
Explanation:

Mid-point of $PR=\left(\cfrac{-2+3}2,\cfrac{2-2}2\right)=\left(\cfrac12,0\right)$

Let coordinate of $S=(x,y)$
Mid-point of $QS=\left(\cfrac{3+x}2,\cfrac{2+y}2\right)$
Mid-point of $PR=$Mid-point of $QS$
$\Rightarrow\left(\cfrac12,0\right)=$$\left(\cfrac{3+x}2,\cfrac{2+y}2\right)$
We have $\cfrac12=\cfrac{3+x}2\Rightarrow x=-2$ and $0=\cfrac{2+y}2\Rightarrow y=-2$
Coordinate of $S=(-2,-2)$
Hence, B is the correct option.

$M$ is the midpoint of the straight line $PQ$. If $P(-2,9)$ and $M$ is $(4,3)$, find the coordinates of $Q$.

  1. $(1,6)$

  2. $(10,-3)$

  3. $(10,6)$

  4. $(8,-3)$


Correct Option: B
Explanation:
$\dfrac{x-2}{2}=4$ and $\dfrac{y+9}{2}=3$
 $\Rightarrow x=10$ and $y=-3$ 
           $a(10,-3)$

$M(2, 6)$ is the midpoint of $\overline {AB}$. If $A$ has coordinates $(10, 12)$, the coordinates of $B$ are

  1. $(6, 10)$

  2. $(-6, 0)$

  3. $(-8, -4)$

  4. $(18, 16)$

  5. $(22, 18)$


Correct Option: B
Explanation:

Mid point $=$ $\dfrac{x _1+x _2}{2}, \dfrac{y _1+y _2}{2}$

Here mid-point $=(2,6)$
$x _1=10,y _1=12$
$\Rightarrow (2,6)=\dfrac{10+x _2}{2},\dfrac{12+y _2}{2}$
$\Rightarrow \dfrac{10+x _2}{2}=2; \dfrac{12+y _2}{2}=6$
$\Rightarrow x _2=4-10;y _2=12-12$
$\Rightarrow x _2=-6,y _2=0$
$\therefore $ co-ordinates of $B=(-6,0)$.

If the mid-point between the points $(a+ b, a- b)$ and $(-a, b)$ lies on the line $ax + by = k$, what is k equal to?

  1. $\dfrac ab$

  2. $a + b$

  3. $ab$

  4. $a - b$


Correct Option: C
Explanation:

Mid point of points $(a+b,a-b),(-a,b)$ is $\left( \dfrac { a+b-a }{ 2 } ,\dfrac { a-b+b }{ 2 }  \right) =\left(\dfrac { b }{ 2 } ,\dfrac { a }{ 2 } \right)$

It lies on line $ax+by=k$ then 
$\Rightarrow a\times \dfrac{b}{2}+b\times \dfrac{a}{2}=k$
$\Rightarrow ab=k$

If a point $C$ be the mid-point of a line segment $AB$, then $AC = BC = (...) AB$.

  1. $3$

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

  3. $2$

  4. $\dfrac{1}{4}$


Correct Option: B
Explanation:

If $C$ is the midpoint of $AB$, then $C$ divides $AB$ in equal segments. Those segments are $AC$ and $AB$.

therefore, $AC=BC$. 
 $AC+BC=AB$ (As AC and BC are the segments of $AB$)
$\Rightarrow 2AC=2BC=AB\ \Rightarrow AC=BC=\dfrac{1}{2} AB$

Say true or false.
The mid-point of the line segment joining the points $P(x _1, y _1)$ and $Q(x _2, y _2)$ is 
$\dfrac {x _1+x _2}{2}, \dfrac {y _1+y _2}{2}.$
  1. True

  2. False


Correct Option: A

If $O(0,0)$ and $P(-8,0)$ then co-ordinates of its midpoint are________.

  1. $(-4,0)$

  2. $(4,0)$

  3. $(0,-4)$

  4. $(0,0)$


Correct Option: A
Explanation:

$O(0,0) $ and $P(-8,0)$ are given points.


The coordinates of the midpoint of 

$\overline{OP}=\left(\dfrac{x _1+x _2}{2},\dfrac{y _1+y _2}{2}\right)$

         $= \left( \dfrac {0-8}{2},\dfrac {0-0}{2}\right)$

         $=(-4,0)$

The mid point of line $AB$ with $A(2,3)$ and $B(5,6)$

  1. $(3.5,4.5)$

  2. $(3,4)$

  3. $(4,5)$

  4. None of these


Correct Option: A
Explanation:

Given points $A(2,3)\equiv(x _1,y _1)$ and $B(5,6)\equiv(x _2,y _2)$


Mid points given as ,


$\Rightarrow\left(\dfrac{x _1+x _2}2,\dfrac{y _1+y _2}2\right)$

$\Rightarrow\left(\dfrac{2+5}2,\dfrac{3+6}2\right)$

$\Rightarrow\left(3.5,4.5\right)$

$(5,-2)$ is the middle line segment joining the parts $\left(\dfrac {x}{2},\dfrac {y+1}{2}\right)$ and $(x+1,y-3)$ then find the value of $x$ & $y$.

  1. True

  2. False


Correct Option: A

Find the area of the triangle formed by joining the mid points of the sides of the triangle whose vertices are $(0.-1), (2, 1) and (0, 3)$

  1. $4$

  2. $8$

  3. $1$

  4. $2$


Correct Option: B
Explanation:

$\\Area\>of\>triangle\>=4\times\>of\>triangle\>formed\>using\>mid-point\>\\=4\times(\frac{1}{2})[x-1(y _2-y _3)+x _2(y _3+y _1)+x _3(y _1-y _2)]\\=2[0+2(3-1)+0]=8sq\>unit$

What is the y intercept of the line that is parallel to $y=3x,$ and which bisects the area of rectangle with corners at $(0,0), (4,0) ,(4,2) $ and $(0,2)$? 

  1. $ -7$

  2. $-6$

  3. $ -5$

  4. $ -4$


Correct Option: C
Explanation:

Rectangle's midpoints are $ = \left( {\frac{{0 + 4}}{2},\frac{{0 + 2}}{2}} \right) = \left( {\frac{4}{2},\frac{2}{2}} \right) = \left( {2,1} \right)$

Slope line y$=$ 3x will be :
${m _1} = 3$
Parallel line will be
${m _2} = 3$
Equation of line passing through (2 , 1)
$y - {y _1} = m(x - {x _1})$
$(y - 1) = 3(x - 2)$
$y - 1 = 3x - 6$
$y = 3x - 5$
Hence on comparing 
Y- intercept $=$ - 5

The mid-point of the line $(a, 2)$ and $(3, 6)$ is $(2, b)$. Find the numerical values of $a$ and $b$.

  1. $a=1$, $b=6$

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

  3. $a=1$, $b=4$

  4. $a=2$, $b=6$


Correct Option: C
Explanation:

Mid-point of $(a,2)$ and $(3,6)$ is $(2,b)$

=>$(2,b)=\left( \cfrac { a+3 }{ 2 } ,\cfrac { 2+6 }{ 2 }  \right) \ =>a=4-3,b=4\ =>a=1,b=4$

If $(3, -4)$ and $(-6, 5)$ are the extremities of a diagonal of a parallelogram and $(2, 1)$ is its third vertex, then its fourth vertex is?

  1. $(-1, 0)$

  2. $(-1, 1)$

  3. $(0, -1)$

  4. $(-5, 0)$


Correct Option: D

If $(6, -3)$ is the one extremity of diameter to the circle $x^{2}+y^{2}-3x+8y-4=0$ then its other extremity is-

  1. $(3/2, -4)$

  2. $(-3, -5)$

  3. $(3, -5)$

  4. $(3, 5)$


Correct Option: B

The length of the median from the vertex A of a triangle whose vertices are $A (-1, 3),$ B $(1, -1)$ and C$(5,1)$ is 

  1. $5$

  2. $4$

  3. $1$

  4. $3$


Correct Option: A
Explanation:
Length of the median from the vertex $A$ of a triangle $\triangle{ABC}$
Let $AD$ be the median.

$\Rightarrow\,D$ is the midpoint of $BC$

Using midpoint formula,$D=\left(\dfrac{1+5}{2},\,\dfrac{-1+1}{2}\right)=\left(3,\,0\right)$

Length of median $=AD=\sqrt{{\left(-1-3\right)}^2{}+{\left(0-3\right)}^{2}}=\sqrt{16+9}=\sqrt{25}=5$units.

The locus of mid points of chords to the circle $x^{2}+y^{2}-8x+6y+20=0$ which are parallel to the line $3x+4y+5=0$ 

  1. $3x+4y-25=0$

  2. $4x+3y+5=0$

  3. $4x-3y-25=0$

  4. $4x-3y+25=0$


Correct Option: A

If $(2, 3), (-4, 5), (1, -2)$ are the midpoints of the sides $\vec{BC}, \vec{CA}, \vec{AB}$ of $\triangle ABC$, then the equation of $\vec{AB}$ is 

  1. $3x-y-5=0$

  2. $x+3y+5=0$

  3. $x+3y-11=0$

  4. $3x-y+17=0$


Correct Option: A

The point on $X-axis$ equidistant from $(2,3)$and $(1,5)$ is

  1. $\left( \dfrac { -13 }{ 2 } ,0 \right) $

  2. $\left( \dfrac { 13 }{ 2 } ,0 \right) $

  3. $(13,0)$

  4. $none\ of\ these$


Correct Option: A

Let ${P} _{1}$ and ${P} _{2}$ be two fixed points in $xy-plane$. A line ${L} _{1}=0$ passes through ${P} _{1}$ intersects $y-axis$ at $B$ and the line ${L} _{2}=0$ passes through ${P} _{2}$ and intersects $x-axis$ at $A$. If ${L} _{1}=0$ and ${L} _{2}=0$ are perpendicular then the locus of mid-point of$AB$ is

  1. $Straight line$

  2. $Circle$

  3. $Ellipse$

  4. $Parabola$


Correct Option: A

The point which is equidistant from the points $(-1,1,3),(2,1,2),(0,5,6)$ and $(3,2,2)$ is

  1. $(-1,3,4)$

  2. $(3,1,4)$

  3. $(1,3,4)$

  4. $(4,1,3)$4


Correct Option: A

The point (5,0) on y-axis is equidistant from (-1,2) and (3,4).

  1. True

  2. False


Correct Option: A

The co-ordinates of the mid point joining the points $(sin^2 \theta, sec^2 \theta )$ and $(cos^2 \theta - tan^2 \theta)$ is

  1. $(2,1)$

  2. $(\dfrac{-1}{2}, \dfrac{1}{2})$

  3. $(1,1)$

  4. $(\dfrac{1}{2}, \dfrac{1}{2})$


Correct Option: A

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

  1. $\left( -6,0 \right )$

  2. $\left( -7,0 \right )$

  3. $\left( 7,0 \right )$

  4. $\left( -5,0 \right )$


Correct Option: A

Let P be the point (1, 0) and Q a point on the curve ${ y }^{ 2 }=8x$. The locus of mid point of PQ is-

  1. ${ y }^{ 2 }-4x+2=0$

  2. ${ y }^{ 2 }+4x+2=0$

  3. ${ x }^{ 2 }+4y+2=0$

  4. ${ x }^{ 2 }-4y+2=0$


Correct Option: A
Explanation:

$P=(1,0), Q=(h,k)$


$k^2=8h$


Let $(\alpha, \beta)$ be the mid-point of PQ.

$\alpha =\dfrac{h+1}{2}, \beta =\dfrac{k+0}{2}$

$2\alpha-1=h, 2\beta=k$

$(2\beta)^2=8(2\alpha-1)$

$\beta^2=4\alpha-2$

$\implies y^2-4x+2=0$

The co -ordinates of the midpoint of a line segment joining $ p(5,7) $ and $ Q (-3,3) $ are........

  1. $ (2,4) $

  2. $ (1,5 ) $

  3. $ (4,2 ) $

  4. $ (2,5 ) $


Correct Option: B
Explanation:
Given,

$P(5,7),Q(-3,3)$

mid point is given by,

$(x,y)=\left ( \dfrac{5-3}{2},\dfrac{7+3}{2} \right )$

$\Rightarrow (x,y)=(1,5)$

If Q is a variable point on $x^2=4y$ and O is the origin, the locus of mid point OQ is equation of 

  1. an ellipse

  2. a parabola

  3. hyperbola

  4. None of these


Correct Option: A

The locus of the mid point of the portion intercepted between the axes by the line $x{\,}cos\alpha+y{\,}sin{\,} \alpha=p$, where $p\inR$, is

  1. $x^2+y^2=\dfrac{4}{p^2}$

  2. $\dfrac{1}{x^2}+\dfrac{1}{y^2}=\dfrac{4}{p^2}$

  3. $\dfrac{1}{x^2}-\dfrac{1}{y^2}=\dfrac{4}{p^2}$

  4. $\dfrac{1}{x^2}+\dfrac{1}{y^2}=\dfrac{2}{p^2}$


Correct Option: A

Locus of the midpoints of the intercepts between the co-ordinate Axes by the lines passing through (a, 0) does not intersect

  1. X axis

  2. Y axis

  3. Y=x

  4. Y=a


Correct Option: A

If the $1st$ point of trisection of AB is $(t, 2t)$ and the ends A, B move on $x$ and $y$ axis respectfully, then the focus of midpoint of AB is 

  1. $x=y$

  2. $2x= y$

  3. $4x= y$

  4. $x= 4y$


Correct Option: C

I every points on the line $(a _{1}-a _{2})x+(b _{1}-b _{2}),y=c$ is equidistance from the points $(a _{1},b _{1})$  and $(a _{2},b _{2})$ then $2c=$  

  1. $a _{1}^{2}-b _{1}^{2}+a _{2}^{2}-b _{2}^{2}$

  2. $a _{1}^{2}+b _{1}^{2}+a _{2}^{2}+b _{2}^{2}$

  3. $a _{1}^{2}+b _{1}^{2}-a _{2}^{2}-b _{2}^{2}$

  4. $None\ of\ these$


Correct Option: B

The line equally inclined to the coordinates axes and equidistant from points A(1, -2) and B(3, 4) is

  1. x+y=2, x+y=3

  2. x-y=3, x-y=1

  3. x-y=1, x+y=3

  4. x+y=2, x-y=3


Correct Option: A

Let $O$ be the origin and $A$ be a point on the curve $y^{2}=4x$. then locus of midpoint of $OA$ is 

  1. $x^{2}=4y$

  2. $x^{2}=4y$

  3. $y^{2}=16x$

  4. $y^{2}=2x$


Correct Option: A

The midpoint of the interval in which $x^{2}-2(\sqrt{-x})^{2}-3<0$ is satisfied, is

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

  2. $-2$

  3. $\dfrac{1}{2}$

  4. $\dfrac{-3}{4}$


Correct Option: A

A tangent to the circle $x^{2}+y^{2}=a^{2}$ meets the axes at points A and B. The locus of the mid point of AB is 

  1. $\frac{1}{x^{2}}+\frac{1}{y^{2}}=\frac{1}{a^{2}}$

  2. $\frac{1}{x^{2}}+\frac{1}{y^{2}}=\frac{4}{a^{2}}$

  3. $\frac{1}{x^{2}}+\frac{1}{y^{2}}=4a^{2}$

  4. $\frac{1}{x^{2}}+\frac{1}{y^{2}}=\frac{a^{2}}{4}$


Correct Option: A

If $A=(1, 2, 3)$ and $B(3, 5, 7)$ and P, Q are the points on AB such that AP$=$PQ$\neq$QB, then the mid point of PQ is?

  1. $(2, 3, 5)$

  2. $\left(2, \dfrac{7}{2}, 5\right)$

  3. $(2, 4, 5)$

  4. $(4, 7, 0)$


Correct Option: A

The locus of the mid-point of a chord of the circle ${ x }^{ 2 }+{ y }^{ 2 }=4$ which subtends a right angle at the origin, is

  1. $x + y = 2$

  2. $x^2 + y^2 = 1$

  3. $x^2 + y^2 = 2$

  4. $x + y = 1$


Correct Option: A

The locus of the mid-point of that chord of parabola which subtends right angle on the vertex will be :

  1. $y^{ { 2 } }-2ax+4{ a }^{ 2 }=0$

  2. $y ^ { 2 } = a ( x - 4 a )$

  3. $y ^ { 2 } = 4 a ( x - 4 a )$

  4. $y^{ { 2 } }+3ax+4{ a }^{ 2 }=0$


Correct Option: A

The locus of the mid-point of that chord of parabola which subtends right angle on the vertex will be

  1. $y ^ { 2 } - 2 a x + 8 a ^ { 2 } = 0$

  2. $y ^ { 2 } = a ( x - 4 a )$

  3. $y ^ { 2 } = 4 a ( x - 4 a )$

  4. $y ^ { 2 } + 3 a x + 4 a ^ { 2 } = 0$


Correct Option: A

The locus of the middle points of chords of length $4$ on the circle $x^ {2}+y^ {2}=16$

  1. A straight line

  2. A circle of radius

  3. A circle of radius $2\sqrt {3}$

  4. An ellipse


Correct Option: A

If the coordinates of the mid-points of side $AB$ and $AC$ of $\triangle ABC$ are $D(3,5)$ and $E(-3,-3)$ respectively, the $BC=$

  1. $10$

  2. $15$

  3. $20$

  4. $30$


Correct Option: A

Find a point on the y-axis which equidistant from the points $A(6,5)$ and $B(-4,3)$

  1. $(0,9)$

  2. $(9,0)$

  3. $(3,0)$

  4. $(4,0)$


Correct Option: A

If $(-6,-4)$ and $(3,5)$ are the extremities of the diagonals of a parallelogram and $(-2,1)$ is its third vertex, then its fourth vertex is 

  1. $(-1,0)$

  2. $(0,-1)$

  3. $(-1,1)$

  4. none of these


Correct Option: A
Explanation:
Given,

$P(-6,-4),Q(3,5),R(-2,1),S(\alpha ,\beta )$

Let $P$ and $Q$ are the extremities of diagonals of a parallelogram, and 

$R$ and $S$ will be the extremities of diagonals of a parallelogram

Now,

midpoint of $PQ=\dfrac{3-6}{2},\dfrac{5-4}{2}=\dfrac{-3}{2},\dfrac{1}{2}$

midpoint of $RS\Rightarrow \dfrac{-2+\alpha }{2}=-\dfrac{3}{2}$

$\Rightarrow \alpha =-3+2=-1$

Now,

$\dfrac{\alpha +\beta }{2}=\dfrac{1}{2}$

$\Rightarrow \beta =0$

Therefore, coordinates of 4th vertex is $(-1,0)$

The mid point of the line joining the points  $\left( \log _ { 2 }  8 , \log _ { 4 }  16 \right)$  and  $\left( \sin 90 ^ { \circ } , \cos \theta \right)$  is

  1. $\sqrt { 2 }$

  2. $\sqrt { 3 }$

  3. $\sqrt { 5 }$

  4. $\sqrt { 7 }$


Correct Option: A

A (a,b) and (0,0) are two fixed points, ${ M } _{ 1 }$ is the mid points of AB, ${ M } _{ 2 }$ is the midpoint of $A{ M } _{ 1 },{ M } _{ 3 }$ is the midpoint of $A{ M } _{ 2 }$ and so on then ${ M } _{ 5 }$ =in

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

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

  3. $\left( \dfrac { 31a }{ 32 } ,\dfrac { 15b }{ 32 } \right) $

  4. $\left( \dfrac { 63a }{ 64 } ,\dfrac { 15b }{ 64 } \right) $


Correct Option: A

If an triangle ABC, A = {1, 10}, circumference = $\left( -\dfrac { 1 }{ 3 } ,\dfrac { 2 }{ 3 }  \right) $ and orthocenter = $\left( \dfrac { 11 }{ 3 } ,\dfrac { 4 }{ 3 }  \right) $ then the co-ordinate of mid-point of side opposite to A is ________.

  1. (1, 11/3)

  2. (1, 5)

  3. (1, -3)

  4. (1, 6)


Correct Option: A

The coordinates of the middle point of the chord of circle ${ x }^{ 2 }+{ y }^{ 2 }-6x=2y-54=0$ which is cut off by the line $2x-5y+18=0$ are __________.

  1. (1,4)

  2. (2,4)

  3. (4,1)

  4. (1,1)


Correct Option: A

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

  1. (-6, 0)

  2. (-7, 0)

  3. (7, 0)

  4. None of these


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?

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


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

  1. $(0.5, 0.8)$

  2. $(-0.5, -0.8)$

  3. $(-0.5, 0.8)$

  4. $(0.5, -0.8)$


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

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

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

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

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


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:

  1. $x= 9, y=5$

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

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

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


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:

  1. $(-7,6)$

  2. $(2, -2)$

  3. $(7,-6)$

  4. $(-2, -2)$


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:

  1. $(-3,2)$

  2. $(2, -3)$

  3. $(3,2)$

  4. $(-2, -3)$


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 $(5,6)$.

  1. $(-1,7)$

  2. $(2,7)$

  3. $(-2,7)$

  4. $(1,7)$


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

  1. $(-7,11)$

  2. $(7,-11)$

  3. $(7,11)$

  4. $(-7,-11)$


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:

  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


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) $

Find the mid-point of AB where A and B are the points $(-5, 11)$ and $(7,3)$, respectively.

  1. $(1,7)$

  2. $(0,0)$

  3. $(1,0)$

  4. $(0,7)$


Correct Option: A
Explanation:

Midpoint 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 AB $= \left( \dfrac { -5 + 7 }{ 2 } ,\dfrac { 11 + 3 }{ 2 } 

\right) = (1, 7) $

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.

  1. $(5,1)$

  2. $(1,1)$

  3. $(1,5)$

  4. $(1,1-)$


Correct Option: B
Explanation:

Let the vertices of the parallelogram be $A (-2,-1), B(1,0),  C(4,3),  D(1,2) $

The diagonals AC and BD would meet at the midpoint of AC and BD.

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) $

Hence, 
mid point of AC $= \left( \cfrac { -2+4 }{ 2 } ,\cfrac { -1+3}{ 2 }  \right) = (1,1) $

The mid-point of a line is $(-4,-2)$ and one end of the line is $(-6,4)$. The co-ordinates of the other end are

  1. $(2,-8)$

  2. $(-2,8)$

  3. $(-2,-8)$

  4. $(2,8)$


Correct Option: C
Explanation:

The co-ordinate of the mid point of a line segment whose end points are $(x _1,y _) $ and $(x _2,y _2)$ are given by $(\cfrac{x _1+x _2}{2},\cfrac{y _1+y _2}{2})$

Substituting $(x _1,y _1)=(-6,4)$ and mid-point $(-4,-2)$.
$\Rightarrow (-4,-2)=(\cfrac{-6+x _2}{2},\cfrac{4+y _2}{2})$
Equating the $x-$coordinates
$\Rightarrow -4=\cfrac{-6+x _2}{2}$
$\Rightarrow -4\times 2=-6+x _2$
$\Rightarrow x _2=6-8$
$\Rightarrow x _2=-2$
Equating the $y-$coordinates

$\Rightarrow -2=\cfrac{4+y _2}{2}$
$\Rightarrow -2\times 2=4+y _2$
$\Rightarrow -y _2=4+4$
$\Rightarrow y _2=-8$

The coordinates of the other end are $(-2,-8)$.

The end points of a diagonal of a parallelogram are $(1, 3)$ and $(5, 7)$, then the mid-point of the other diagonal is ..........

  1. $(1, 7)$

  2. $(3, 5)$

  3. $(5, 3)$

  4. $(7, 1)$


Correct Option: B
Explanation:

Given end points of diagonal of parallelogram $(1,3)$ and $(5,7)$

We know that the mid point of two point $(x _1,y _1) \ and \  (x _2,y _2)$ are
$\dfrac{x _{1}+x _{2}}{2},\dfrac{y _{1}+y _{2}}{2}$
$\Rightarrow \dfrac{1+5}{2},\dfrac{3+7}{2}$
$\Rightarrow( 3,5)$

Calculate mid point of $A(5,\,3)$ and $B(9,\,8)$

  1. $\dfrac{11}{2},\,7$

  2. $7,\,\dfrac{11}{2}$

  3. $7,\,11$

  4. $14,\,11$


Correct Option: B
Explanation:

The mid point of line segment joining $(x _1,\,y _1)$ and $(x _2,\,y _2)$ is $\begin{pmatrix}\dfrac{x _1+x _2}{2},\,\dfrac{y _1+y _2}{2}\end{pmatrix}$
Mid point of AB $\Rightarrow\begin{pmatrix}\dfrac{5+9}{2},\,\dfrac{3+8}{2}\end{pmatrix}$
$\Rightarrow\begin{pmatrix}7,\,\dfrac{11}{2}\end{pmatrix}$

Three vertices of rhombus taken in order are $(2, -1), (3, 4)$ and $(-2, 3)$. Find the fourth vertex.

  1. $(1, 2)$

  2. $(-3, -2)$

  3. $(3, 2)$

  4. None of these


Correct Option: B
Explanation:

Let M be the mid point of one diagonal formed by (2, -1) and (-2, 3) then using mid-point method;
Coordinate of $M \left (\dfrac {2 - 2}{2}, \dfrac {-1 + 3}{2}\right ) = (0, 1)$
$\Rightarrow \left (\dfrac {x + 3}{2}, \dfrac {y + 4}{2}\right ) = (0, 1)$
$\Rightarrow x = -3$ and $y = -2$
$\therefore (-3, -2)$ is coordinate of $D$

What is the midpoints between the coordinates $(-1, 2)$ and $(-1, -6)$?

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

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

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

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


Correct Option: C
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
$=$ $\left(\dfrac{-1-1}{2},\dfrac{2-6}{2}\right)$
$=$ $\left(\dfrac{-2}{2},\dfrac{-4}{2}\right)$
Therefore, midpoint is $\left(-1, -2\right)$

What is the midpoints between the coordinates $(0, -6)$ and $(4, -4)$?

  1. $\left(-2, -5\right)$

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

  3. $\left(2, -4\right)$

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


Correct Option: D
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
$=$ $\left(\dfrac{0+4}{2},\dfrac{-6-4}{2}\right)$
$=$ $\left(\dfrac{4}{2},\dfrac{-10}{2}\right)$
Therefore, midpoint is $\left(2, -5\right)$.

Find the centre of circle, if the coordinates of two ends of diameter are $(-1, 7)$ and $(11, 5)$

  1. $(5, 6)$

  2. $(-3, 2)$

  3. $(10, 12)$

  4. $(12, 2)$


Correct Option: A
Explanation:

$x = \dfrac {11 - 1}{2} = 5$ and $y = \left (\dfrac {5 + 7}{2}\right )$ $= 6$ (hint: using mid-point formula for centre).
$\therefore (5, 6)$ is coordinate of centre

Find the midpoint between the coordinates $(9, 3)$ and $(1, 1)$.

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

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

  3. $\left(5, 1\right)$

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


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
$=$ $\left(\dfrac{9+1}{2},\dfrac{3+1}{2}\right)$
$=$ $\left(\dfrac{10}{2},\dfrac{4}{2}\right)$
Therefore, midpoint is $\left(5, 2\right)$

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 $

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

If $(-2,3), (4,-3), (4,5)$ are mid-points of the sides of a triangle, find the coordinates of the centroid of the triangle formed by these mid-points.

  1. $\left (3,\dfrac43 \right )$

  2. $\left (2,\dfrac43 \right )$

  3. $\left (2,\dfrac53 \right )$

  4. $\left (3,\dfrac53\right )$


Correct Option: C
Explanation:

Let the given vertices of a triangle be A$(-2,3)$ and B $(4,-3)$ let the third vertex be C $(4,5)$.


Let Centroid be $G= \left [\dfrac {x _1 + x _2 + x _3}{3} , \dfrac{y _1 +y _2 + y _3}{3}\right ]$


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

$\Rightarrow G$ = $\left (\dfrac{6 }{3} , \dfrac{5}{3}\right )$

$\Rightarrow G$ = $\left (2 , \dfrac {5}{3}\right )$.

Find the third vertex of a triangle, if two of its vertices are $(-3,1), (0,-2)$ and centroid is at the origin.

  1. $(3,4)$

  2. $(2,1)$

  3. $(3,2)$

  4. $(3,1)$


Correct Option: D
Explanation:

Let the third vertex be $(a,b)\equiv(x _1,y _1),(-3,1)\equiv(x _2,y _2), (0,-2)\equiv(x _3,y _3)$


Centroid of triangle is $(0,0)$

$G =\left [\dfrac {x _1 + x _2 + x _3}{3} , \dfrac{y _1 +y _2 + y _3}{3}\right ]$

$\Rightarrow \left( \dfrac { a-3+0 }{ 3 } ,\dfrac { b+1-2 }{ 3 }  \right) =(0,0)$

$\Rightarrow\dfrac{a-3}3=0$ and $\dfrac{b-1}3=0$

$\Rightarrow a=3,b=1$

So the third vertex is $(3,1)$

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

  1. $(-2,1)$

  2. $(2,1)$

  3. $(-2,-1)$

  4. None of these


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.

  1. $(-12,8)$

  2. $(-12,-8)$

  3. $(12,8)$

  4. None of these


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).

  1. $(5,6)$

  2. $(6,6)$

  3. $(4,4)$

  4. None of the above


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 $(2,-6)$.

  1. $(3,4)$

  2. $(2,-2)$

  3. $(3,0)$

  4. None of the above


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

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

  2. $a = b = 2$

  3. $a= 1 = b$

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


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.

  1. $(1,1)$

  2. $(1,3)$

  3. $(5,1)$

  4. None of these


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 ?

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

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

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

  4. $(1,4)$


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 

  1. $(-9,7)$

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

  3. $(9,-7)$

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


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

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

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

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

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


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:

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

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

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

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


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) $


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