Tag: distances and midpoints

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Questions Related to distances and midpoints

Multiple choice maths fundamentals pair of straight lines distances and midpoints distance formula in 2d

Find $a$ if the distance between $(a , 2)$ and $(3 , 4)$  is $8 $

  1. $ 3 \, \pm \, \sqrt {60}$
  2. $ 4 \, \pm \, \sqrt {60}$
  3. $ 3 \, \pm \, \sqrt {6}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

By square of distance formual:


$8^2=(3-a)^2+(4-2)^2$

$=>64=a^2-6a+13$

$=>a^2-6a-51=0$

solving the quadratic we get:

$a=(3+\sqrt(60)$ or $(3-\sqrt(60))$.

Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

The equation of the line which passes through $(0,0)$ and $(1,1)$ is ____________

  1. $y=x$
  2. $y=-x$
  3. $y=1$
  4. $x=1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation of the line which passes through (0,0) and (1,1) is y=x

As we know, the two points are $(0,0) (1,1)$
slope $ m$ $=\dfrac { { y } _{ 2 }-{ y } _{ 1 } }{ { x } _{ 2 }-{ x } _{ 1 } } $  $=\dfrac { 1-0 }{ 1-0 } =1$
Standard equation of the line is $y-{ y } _{ 1 }=m\left( x-{ x } _{ 1 } \right) $
Substituting value of $m =1 $ and point $(x,y) =  (0,0)$ we get:
$y-0=1\left( x-0 \right) $
 $y=x$ is the required equation of the line

Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

The equation of a straight line passing through points $(0,0)$ and $(1,5)$ is given by:

  1. $y=x$
  2. $y=5x$
  3. $5y=x$
  4. $y=x+5$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given a line passes through $A(0,0)$ & $B(1,5)$ 

We know that the equation of line which passes through $(a _1,b _1)$ & $\left( { a } _{ 2 },{ b } _{ 2 } \right) $ is 
$y-{ b } _{ 1 }=\cfrac { { b } _{ 2 }-{ b } _{ 1 } }{ { a } _{ 2 }-{ a } _{ 1 } } (x-{ a } _{ 1 })\ \therefore \quad y-0=\cfrac { 5-0 }{ 1-0 } (x-0)\ y=5x$ 

Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

For any real value of $\lambda$, the equation $2x^2+3y^2-8x-6y+11-\lambda =0$ doesn't represents a pair of straight lines?

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The second degree terms cannot be factorized into two linear factors or else $h^2-ab=0-6=-$ve. Hence the given equation does not represent a pair of lines whatever $\lambda$ may be.

Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

The base at a triangle passes through a fixed point $(a, b)$ and its sides are respectively bisected at right angles by the lines $y^{2} - 4xy - 5x^{2} = 0$. Find locus of its vertex.

  1. $2 \, (x^2 \, + \, y^2) + (3a + 2b) x + (2a - 3b) y = 0$
  2. $2 \, (x^2 \, + \, y^2) - (3a + 2b) x + (2a - 3b) y = 0$
  3. $2 \, (x^2 \, + \, y^2) + (3a + 2b) x - (2a - 3b) y = 0$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The lines are y^2 - 4xy - 5x^2 = 0, which factor into (y - 5x)(y + x) = 0. Using the properties of the orthocenter and the given fixed point, the locus of the vertex is derived as 2(x^2 + y^2) + (3a + 2b)x + (2a - 3b)y = 0.

Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

The differential equation of the curve such that the ordinates of any point is equal to the corresponding subnormal at that point is

  1. a linear equation

  2. a non-homogeneous equation

  3. an equation with separable variable

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The differential equation of the curve such that the ordinate of the any point is equal to the  corresponding subnormal at that point is a linear equation.

Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

Let a and b be non-zero real numbers. Then, the equation $(ax^2+by^2+x)(x^2-5xy+6y^2)=0$ represents.

  1. Four straight lines, when $c=0$ and a, b are of the same sign
  2. Two straight lines and a circle, when $a=b$, and c is of sign opposite to that of a

  3. Two straight lines and a hyperbola, when a and b are of the same sign and c is of sign opposite to that of a

  4. A circle and an ellipse, when a and b are of the same sign and c is of a sign opposite to that of a
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation (ax^2 + by^2 + x)(x^2 - 5xy + 6y^2) = 0 represents a conic and two lines. If a = b, the first part represents a circle. If c (the constant term in the conic) is of opposite sign to a, it confirms the conic properties.

Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

The equation $x^2y^2-2xy^2-3y^2-4x^2y+8xy+12y=0$ represents.

  1. A pair of lines

  2. Pair of lines and a circle

  3. A pair of lines and a parabola

  4. Four lines forming a square
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Collecting the terms $y^2$ and y the given equation can be written as
$y^2(x^2-2x-3)-4y(x^2-2x-3)=0$
or $(x-3)(x+1)y(y-4)=0$
It represents four lines $x=-1$, $x=3$, $y=0$ and $y=4$.
These two sets of parallel lines form a square of side four.

Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

Perpendicular distance between line $2x + y  =5,  2x + y  =3$ 

  1. $\dfrac{1}{{\sqrt 2 }}$
  2. $\sqrt 2 $
  3. $\dfrac{{2 }}{\sqrt 5}$
  4. $\dfrac{3}{{\sqrt 2 }}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The given lines are


$2x+y=3\cdots(1)$

$2x+y=5\cdots(2)$

The perpendicular distance between lines is given as 

$\dfrac{|c _1-c _2|}{\sqrt {a^2+b^2}}$

$\dfrac{|5-3|}{\sqrt{ 2^2+1^2}}$

$\dfrac{2}{\sqrt 5}$