Tag: linear graphs

Questions Related to linear graphs

The abscissa of two points A and B are the roots of the equation ${x^2} + 2ax - {b^2}$ and their ordinates are the root of the equation ${x^2} + 2px - {q^2}=0$. the equation of the circle with AB as diameter is 

maths linear graphs quadrants the cartesian plane the cartesian system

  1. ${x^2} + {y^2} + 2ax + 2py + {b^2} + {q^2} = 0$

  2. ${x^2} + {y^2} - 2ax - 2py - {b^2} - {q^2} = 0$

  3. ${x^2} + {y^2} + 2ax + 2py - {b^2} - {q^2} = 0$

  4. None of these

Show answer
Correct Option: C
Explanation:
${ x }^{ 2 }+2ax-{ b }^{ 2 }=0$

Let roots be ${ x } _{ 1 }$ & ${ x } _{ 2 }$

${ x } _{ 1 }+{ x } _{ 2 }=\dfrac{-b}{a}=-2a$

$ { x } _{ 1 }{ x } _{ 2 }=\dfrac{c}{a}=-{ b }^{ 2 }$

$ { x }^{ 2 }+2px-{ q }^{ 2 }=0$

Let roots be ${ y } _{ 1 }$ & ${y } _{ 2 }$

${ y } _{ 1 }+{ y } _{ 2 }=-2p$

$ { y} _{ 1 }{ y } _{ 2 }=-{ q }^{ 2 }$

Equation in diametric form is 
${ x }^{ 2 }+{ y }^{ 2 }-({ x } _{ 1 }+{ x } _{ 2 })x-({ y } _{ 1 }+{ y } _{ 2 })y+{ x } _{ 1 }{ x } _{ 2 }+{ y } _{ 1 }{ y } _{ 2 }=0$

$ { x }^{ 2 }+{ y }^{ 2 }+2ax+2py-{ b }^{ 2 }-{ q }^{ 2 }=0$

The acute angle between the lines $x-y=0$ and $y=0$ is

maths linear graphs quadrants the cartesian plane the cartesian system

  1. $30^{\circ}$

  2. $45^{\circ}$

  3. $60^{\circ}$

  4. $75^{\circ}$

Show answer
Correct Option: B
Explanation:

Given 


$x-y=0$

$y=0$ represents $x-axis$ 

So the slope of line $x-y=0$ is $-\dfrac 1{-1}=1$

$\implies \tan \theta =1$

$\tan \theta =\tan \dfrac \pi 4$

$\theta =\dfrac \pi 4$

$\theta =45^{\circ}$

 If the distance between the points $\left( {a\,\cos {{48}^ \circ },0} \right)$ and $\left( {\,0,a\,\cos {{12}^ \circ }} \right)$ is d,then ${d^2} - {a^2} = $

maths linear graphs quadrants the cartesian plane the cartesian system

  1. ${a^2}\left( {\sqrt 5 - 1} \right),/4$

  2. ${a^2}\left( {\sqrt 5 + 1} \right),/4$

  3. ${a^2}\left( {\sqrt 5 - 1} \right),/8$

  4. $\dfrac{a^2( {\sqrt 5 + 1} )}{8}$

Show answer
Correct Option: D
Explanation:
Using distance formula,${d}^{2}={\left(a\cos{{48}^{\circ}}-0\right)}^{2}+{\left(0-a\cos{{12}^{\circ}}\right)}^{2}$

${d}^{2}-{a}^{2}={\left(a\cos{{48}^{\circ}}-0\right)}^{2}+{\left(0-a\cos{{12}^{\circ}}\right)}^{2}-{a}^{2}$

$={a}^{2}{\cos}^{2}{{48}^{\circ}}+{a}^{2}{\cos}^{2}{{12}^{\circ}}-{a}^{2}$

$={a}^{2}{\cos}^{2}{{48}^{\circ}}-{a}^{2}+{a}^{2}{\cos}^{2}{{12}^{\circ}}$

$={a}^{2}{\cos}^{2}{{48}^{\circ}}-{a}^{2}\left(1-{\cos}^{2}{{12}^{\circ}}\right)$

$={a}^{2}{\cos}^{2}{{48}^{\circ}}-{a}^{2}{\sin}^{2}{{12}^{\circ}}$

We know that ${\cos}^{2}{B}-{\sin}^{2}{A}=\cos{\left(A+B\right)}\cos{\left(A-B\right)}$

$={a}^{2}\cos{\left({12}^{\circ}+{48}^{\circ}\right)}\cos{\left({12}^{\circ}-{48}^{\circ}\right)}$

$={a}^{2}\cos{{60}^{\circ}}\cos{\left(-{36}^{\circ}\right)}$

$={a}^{2}\cos{{60}^{\circ}}\cos{{36}^{\circ}}$ since $\cos{\left(-\theta\right)}=\cos{\theta}$

$={a}^{2}\times\dfrac{1}{2}\times\dfrac{\sqrt{5}+1}{4}$

$=\dfrac{{a}^{2}\left(\sqrt{5}+1\right)}{8}$

If the points $A ( 2,1,1 ) , B ( 0 , - 1,4 ) , C ( K , 3 , - 2 )$ are collinear then $K =$

maths linear graphs quadrants the cartesian plane the cartesian system

  1. k=5

  2. k=4

  3. k=9

  4. k=10

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Correct Option: B

Find the number of points on the straight line which joins $\left( { - 4,\,11} \right)$ to $\left( { 16,\,- 1} \right)$ whose co-ordinates are positive integer.

maths linear graphs quadrants the cartesian plane the cartesian system

  1. $1$

  2. $2$

  3. $3$

  4. $4$

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Correct Option: C

If the points $( 2,0 ) , ( 0,1 ) , ( 4,5 ) \text { and } ( 0 , c )$ are concyclic then the value of $c$ is 

maths linear graphs quadrants the cartesian plane the cartesian system

  1. $1,\dfrac{13}{3}$

  2. $5,\dfrac { 14 } { 3 }$

  3. $5,\dfrac{15}{4}$

  4. none of these

Show answer
Correct Option: A
Explanation:
Consider equation of circle to be $x^2+y^2+2gx+2fg+c=0$
Substituting given points
$4+4g+c=0$ ………….(i)
$1+2f+c=0$ ………..(ii)
$39+8g+10f+c=0$ ……….(iii)
solving them $g=-25/12$; $f=-8/3$
$c=13/3$
$x^2+y^2-\dfrac{25}{6}x-\dfrac{16}{3}y+\dfrac{13}{3}=0$ is equation
Substituting $(0, c)$
$c^2-\dfrac{16}{3}c+13/3=0$
$c=1, 13/3$.

If points $( - 7,5 ) \text { and } \left( \alpha , \alpha ^ { 2 } \right)$ lie on the opposite sides of the line $5 x - 6 y - 1 = 0$ then 

maths linear graphs quadrants the cartesian plane the cartesian system

  1. $\alpha \in [ 0,1 ]$

  2. $a \in [ - 1,0 ]$

  3. $\alpha \in \left( \dfrac { 1 } { 3 } , \dfrac { 1 } { 2 } \right)$

  4. $\alpha \in [ 2,4 ]$

Show answer
Correct Option: C
Explanation:
$(-7, 5)$ & $(\alpha, \alpha^2)$ lie on opposite side of $5x-6y-1=0$

$(-7, 5)\rightarrow 5x-6y-1=5(-7)-6(5)-1$

$=-35-30-1$

$=-66 < 0$

as $(\alpha,\alpha^2)$ should be on the opposite of $5x-6y-1=0$

$\Rightarrow (\alpha, \alpha^2)\rightarrow 5x-6y-1=5\alpha-6\alpha^2-1 > 0$

$\Rightarrow 6\alpha^2-5\alpha +1 < 0$

$\Rightarrow 6\alpha^2-2\alpha -3\alpha +1 < 0$

$\Rightarrow 2\alpha(3\alpha -1)-1(3\alpha -1) < 0$

$\Rightarrow (2\alpha -1)(3\alpha -1) < 0$

$\Rightarrow \dfrac{1}{3} < \alpha < \dfrac{1}{2}$.

If the three distinct points $\left( t,2at+{ at }^{ 3 } \right)$ for $i=1,2,3$are collinear then the sum of the abscissa of the  _________.

maths linear graphs quadrants the cartesian plane the cartesian system

  1. -1

  2. 0

  3. 1

  4. 3

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Correct Option: B

The abscissa of a point on the curve $xy=(a+x)^{2}$, the normal cuts off numerically equal intercepts from the coordinate axes, is

maths linear graphs quadrants the cartesian plane the cartesian system

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

  2. $\sqrt{2}a$

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

  4. $-\sqrt{2}a$

Show answer
Correct Option: A,C
Explanation:
Given,

$xy=(a+x)^2$

differentiating the above equation, we get,

$\Rightarrow y+xy'=2(a+x)$

substituting and solving the above equation, we get,

$\therefore y'=\pm 1$

$y \pm x =2(a+x)$

$\dfrac{(a+x)^2}{x}\pm x=2(a+x)$

$\Rightarrow \pm x=2(a+x)-\dfrac{(a+x)^2}{x}$

$\pm x^2=(2+x)[x-a]$

$\pm x^2=x62-a^2$

$\Rightarrow 2x^2=a^2$

$\therefore x=\pm \dfrac{a}{\sqrt 2}$

To remove Xy term from the second degree equation $5x^2 + 8xy + 5y^2 + 3x + 2y + 5 = 0$, the coordinates axes are rotated through an angle q, then q equals.

maths linear graphs quadrants the cartesian plane the cartesian system

  1. $\pi/2$

  2. $\pi/4$

  3. $\pi/8$

  4. $\pi/8$

Show answer
Correct Option: A