Tag: maths

Questions Related to maths

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