Tag: distances and midpoints

Questions Related to distances and midpoints

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

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

  1. $ 3 \, \pm \, \sqrt {60}$

  2. $ 4 \, \pm \, \sqrt {60}$

  3. $ 3 \, \pm \, \sqrt {6}$

  4. None of these

Show answer
Correct Option: A
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 $(3-\sqrt(60))$.

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

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

  1. $y=x$

  2. $y=-x$

  3. $y=1$

  4. $x=1$

Show answer
Correct Option: A
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

Write an equation of the horizontal line through the point $(7,-5)$

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

  1. $-5y=7x+3$

  2. $y=-5$

  3. $x=7$

  4. none of these

Show answer
Correct Option: B
Explanation:

The equation of a straight line is $y = mx +b$

For a horizontal line $m = 0$.
So, $y = b$ and since the line must pass through $x = 7, y = - 5$, 

$b$ must be $-5$
Therefore, $y=-5$.

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

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

  1. $y=x$

  2. $y=5x$

  3. $5y=x$

  4. $y=x+5$

Show answer
Correct Option: B
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 
$y-{ b } _{ 1 }=\cfrac { { b } _{ 2 }-{ b } _{ 1 } }{ { a } _{ 2 }-{ a } _{ 1 } } (x-{ a } _{ 1 })\ \therefore \quad y-0=\cfrac { 5-0 }{ 1-0 } (x-0)\ y=5x$ 

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?

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

  1. True

  2. False

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Correct Option: A
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.

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.

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

  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

Show answer
Correct Option: A

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

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

  1. a linear equation

  2. a non-homogeneous equation

  3. an equation with separable variable

  4. none of these

Show answer
Correct Option: A
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.

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

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

  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

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Correct Option: B

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

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

  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

Show answer
Correct Option: D
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.

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

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

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

  2. $\sqrt 2 $

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

  4. $\dfrac{3}{{\sqrt 2 }}$

Show answer
Correct Option: C
Explanation:

The given lines are


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

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

The perpendicular distance between lines is given as 

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

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

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