Tag: maths

Questions Related to maths

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

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

  2. $y=-5$

  3. $x=7$

  4. none of these


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:

  1. $y=x$

  2. $y=5x$

  3. $5y=x$

  4. $y=x+5$


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?

  1. True

  2. False


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.

  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


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

  1. a linear equation

  2. a non-homogeneous equation

  3. an equation with separable variable

  4. none of these


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.

  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


Correct Option: B

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


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 $2x + y  =5,  2x + y  =3$ 

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

  2. $\sqrt 2 $

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

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


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 

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

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

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

Equation $4x^{2}+4xy-y^{2}-6x-3y-4=0$ represents a pair of parallel lines, then distance between these lines is

  1. $2\sqrt{5}$

  2. $\sqrt{5}$

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

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


Correct Option: A

The difference of the slopes of the lines $x ^ { 2 } \left( \sec ^ { 2 } \theta - \sin ^ { 2 } \theta \right) - ( 2 \tan \theta ) x y + y ^ { 2 } \sin ^ { 2 } \theta = 0$

  1. $1$

  2. $2$

  3. $3$

  4. $4$


Correct Option: B
Explanation:

According to the question..........

$\begin{array}{l} Let,\, { m _{ 1\,  } }& \, { m _{ 2 } } \ sum\, of\, the\, slope:\, { m _{ 1 } }+{ m _{ 2 } }=\dfrac { { -2h } }{ b } ----(i) \ and,\,  \ product\, of\, slope:{ m _{ 1 } }.\, { m _{ 2 } }=\dfrac { a }{ b } -----(ii) \ Here, \ a{ x^{ 2 } }+2hxy+b{ y^{ 2 } }=0.........(general\, equ\, of\, straight\, line.) \ cofficient\, of: \ a={ \sec ^{ 2 }  }\theta -{ \sin ^{ 2 }  }\theta  \ h=-\tan  \theta  \ b={ \sin ^{ 2 }  }\theta  \ Now,\, value\, put\, { { into } } \ sum\, of\, the\, slope:\, { m _{ 1 } }+{ m _{ 2 } }=\dfrac { { -2h } }{ b } ----(i) \ \Rightarrow { m _{ 1 } }+{ m _{ 2 } }=\dfrac { { -2(-tan\theta ) } }{ { { { \sin   }^{ 2 } }\theta  } } =\dfrac { { 2\sin  \theta \times 2 } }{ { 2{ { \sin   }^{ 2 } }\theta \, .\, \cos  \theta  } } =\dfrac { 4 }{ { 2sin\theta \cos  \theta  } } =\dfrac { 4 }{ { \sin  2\theta  } }  \ and, \ product\, of\, slope:{ m _{ 1 } }+{ m _{ 2 } }=\dfrac { a }{ b } -----(ii) \ \Rightarrow { m _{ 1 } }.\, { m _{ 2 } }=\dfrac { { { { \sec   }^{ 2 } }\theta -{ { \sin   }^{ 2 } }\theta  } }{ { { { \sin   }^{ 2 } }\theta  } } =\dfrac { 1 }{ { { { \sin   }^{ 2 } }\theta \, .\, { { \cos   }^{ 2 } }\theta  } } -1\, \, \, \, \, \, \, \, \left[ { divide\, by\, 4 } \right.  \ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\dfrac { 4 }{ { 4{ { \sin   }^{ 2 } }\theta \, .\, { { \cos   }^{ 2 } }\theta  } } -1\, \, =\dfrac { 4 }{ { { { (\sin  2\theta ) }^{ 2 } }\,  } } -1\,  \ \, \, \, Now,find\, difference: \ \, \, \, \, \, \, \, \, \, { ({ m _{ 1 } }-{ m _{ 2 } })^{ 2 } }={ ({ m _{ 1 } }+{ m _{ 2 } })^{ 2 } }-4{ m _{ 1 } }.\, { m _{ 2 } } \ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, ={ \left( { \dfrac { 4 }{ { \sin  2\theta  } }  } \right) ^{ 2 } }-4\left( { \dfrac { 4 }{ { ({ { \sin   }^{ 2 } }2\theta )\,  } } -1\,  } \right)  \ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\dfrac { { 16 } }{ { ({ { \sin   }^{ 2 } }2\theta )\,  } } -\, \dfrac { { 16 } }{ { ({ { \sin   }^{ 2 } }2\theta )\,  } } +4 \ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \Rightarrow \, \, { ({ m _{ 1 } }-{ m _{ 2 } })^{ 2 } }\, \, \, =4 \ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \Rightarrow \, ({ m _{ 1 } }-{ m _{ 2 } })=+\sqrt { 4 } =2 \ \, \, \, \therefore \, \, \, the\, \, differece\, of\, slope\, \, is\, 2. \ So,\, that\, the\, correct\, option\, is\, B.\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \,  \end{array}$