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Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

In the xy-plane, the parabola with equation $y = (x - 11)^{2}$ intersects the line with equation $y = 25$ at two points, $A$ and $B$. What is the length of $\overline {AB}$?

  1. $10$
  2. $12$
  3. $14$
  4. $16$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given $y=(x-11)^2$

$\Rightarrow  y=x^2-22x+121$
After substitute $ y=25$. we get,
$x^2-22x+121-25=0$
$\Rightarrow x^2-22x+96=0$
$\Rightarrow (x-16)(x-6)=0$
$\Rightarrow  x=16 , x=6$
$\therefore$ two points are  $A, B$ are $(16,25), (6,25)$.
Distance between A and B is 
$\sqrt {(16-6)^2-(25-25)^2}=10$
Hence, option A is correct.

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

Let $y=f(x)$ and $y=g(x)$ be the pair of curves such that
(i) The tangents at point with equal abscissae intersect on y-axis.
(ii) The normal drawn at points with equal abscissae intersect on x-axis and
(iii) curve f(x) passes through $(1, 1)$ and $g(x)$ passes through $(2, 3)$ then the value of $\displaystyle\int^2 _1(g(x)-f(x))dx$ is?

  1. $2$
  2. $3$
  3. $4$
  4. $5$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$y=f(x)$ and $y=g(x)$
(i)
$\dfrac{dy}{dy}=c\Rightarrow c _{1}=1$
for second curve 
$\dfrac{dy}{dy}=c _{2}\Rightarrow c _{2}=1$

(ii)
$\dfrac{dy}{dx}=d _{1}---(1)$
for second curve 
$\dfrac{dy}{dx}=d _{2}----(2)$

(iii)
From eq (1) and (2)
$d _{1}=2=d _{2}$

$\int _{1}^{2} (g(x)-f(x))d(x)$

$\int _{1}^{2} (4-2)d(x)$

$2\int _{1}^{2} d(x)$

$2(2-1)=2$
Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

The number of values of $C$ for which the line $y = 4x + c$ touch the curve $\dfrac {x^{2}}{4} + y^{2} = 1$.

  1. $0$
  2. $1$
  3. $2$
  4. $\infty$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The given curve $\dfrac {x^{2}}{4} + y^{2} = 1$ is an ellipse

Here, $a=2$ and $b=1$

And equation of line is slope form is $y=mx+c$
Also, $c = \sqrt {a^{2}m^{2} + b^{2}}$
$\Rightarrow c = \sqrt {4(16) + 1} = \pm \sqrt {65}$
Hence, $c$ has two values.

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

The point of intersection of line $\dfrac {x - 6}{-1} = \dfrac {y + 1}{0} = \dfrac {z + 3}{4}$ and plane $x + y - z = 3$ is

  1. $(2, 1, 0)$
  2. $(7, -1, -7)$
  3. $(1, 2, -6)$
  4. $(5, -1, 1)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The given line is $\dfrac {x - 6}{-1} = \dfrac {y + 1}{0} = \dfrac {z + 3}{4}=r$(say) ..... $(i)$

And Plane is $x + y - z = 3$ ........ $(ii)$
$\Rightarrow x=-r+6, y=-1, z=4r-3$
Then, the point $(-r + 6, - 1, 4r - 3)$ lies on the line $(i)$. 

It is given that the plane and the line intersects
Thus, the point $(-r + 6, - 1, 4r - 3)$ satisfies the plane
$\Rightarrow (-r + 6) - 1 - (4r - 3) = 3\Rightarrow r = 1$
$\therefore$ Required intersection point $= (5, -1, 1)$.

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

The value that m can take so that the straight line $y=4x+m$ touches the curve $x^{2}+4y^{2}=4$ is 

  1. $\underline{+}\sqrt{45}$
  2. $\underline{+}\sqrt{60}$
  3. $\underline{+}\sqrt{65}$
  4. $\underline{+}\sqrt{72}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation


$C=\underline{+}\sqrt{a^{2}\, m^{2}+b^{2}}$

$={+\sqrt{4(16)+1}}=\underline{+}\sqrt{65}$

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

Find the point of intersection and the inclination of the two lines $Ax+By=A+B$ and $A(x-y)+B(x + y)=2B$.

  1. $(1,1); 45^0$
  2. $(1,2), 60^0$
  3. $(2,1), 75^0$

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

$Ax+By=A+B$    ........(i)

$  A(x-y)+B(x+y)=2B$

$\Rightarrow  (A+B)x+(B-A)y=2B$    .......(ii)

$\Rightarrow  (A+B)x=2B-(B-A)y\\ \Rightarrow x=\dfrac { 2B-(B-A)y }{ A+B } $

Substituting $x$ in (i), we get

$A\left( \dfrac { 2B-(B-A)y }{ A+B }  \right) +By=A+B$

$\Rightarrow \dfrac { 2AB-ABy+{ A }^{ 2 }y+{ B }^{ 2 }y+ABy }{ A+B } =A+B$

$\Rightarrow 2AB-ABy+{ B }^{ 2 }y+{ A }^{ 2 }y+ABy={ A }^{ 2 }+{ B }^{ 2 }+2AB$ 

$\Rightarrow (A^{2}+{ B }^{ 2 })y=A^{2}+{ B }^{ 2 }$

$\Rightarrow y=1 $

Substituting $y$ in $(i)$

$\Rightarrow  Ax+B\left( 1  \right) =A+B\\ \Rightarrow Ax+B=A+B\\ \Rightarrow Ax=A\\ \Rightarrow x=1 $

So, the point of intersection is $(1,1) $.

Slope of (i), ${ m } _{ 1 }=-\dfrac { A }{ B } $.

Slope of (ii), ${ m } _{ 2 }=-\dfrac { (A+B) }{ B-A } =\dfrac { A+B }{ A-B } $

$\tan { \theta  } =\dfrac { { m } _{ 1 }-{ m } _{ 2 } }{ 1+{ m } _{ 1 }{ m } _{ 2 } } \\ \Rightarrow \tan { \theta  } =\dfrac { -\dfrac{A}{B}-\dfrac { A+B }{ A-B }  }{ 1-\dfrac{A}{B}\times\dfrac { A+B }{ A-B }  } $

$ \tan { \theta  } =-\dfrac { \left\{ \dfrac { { A }^{ 2 }+AB-AB+{ B }^{ 2 } }{ B(A-B) }  \right\}  }{ \left\{ \dfrac { { -B }^{ 2 }+AB-{ A }^{ 2 }-AB }{ B(A-B) }  \right\}  } \\ \tan { \theta  } =-\dfrac { { A }^{ 2 }+{ B }^{ 2 } }{ { -(A }^{ 2 }+{ B }^{ 2 }) } =1$

$\Rightarrow \theta=45^{o}$

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

The equation $x-y = 4$ and $x^2 + 4xy + y^2 = 0$ represent the sides of

  1. an equilateral triangle

  2. a right angled triangle

  3. an isosceles triangle

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

$(y+m _{1}x)(y+m _{2}x)=0$


$y^{2}+(m _{1}+m _{2})xy+m _{2}m _{1}x^2=0$

Comparing coefficients we get

$m _{1}+m _{2}=4$

$m _{1}m _{2}=1$

This implies

$m _{1}^{2}+1=4m _{1}$

$m _{1}^{2}-4m _{1}+1=0$

Therefore $m _{1}=2-\sqrt{3}$ and $m _{1}=2+\sqrt{3}$

$\tan A=2-\sqrt{3}$ and $\tan A=2+\sqrt{3}$

Hence $A=15^0$ and $A=75^0$

Corresponding values of $m _{2}=75^{\circ}$ and $15^{\circ}$

Therefore the angle between the lines is $75^0-15^0$
$=60^0$

The equation of the angle bisectors of the lines is $x=\pm y$

The line $x=-y$ is perpendicular to $x-y=4$

Hence the above is an isosceles triangle with the vertical angle being $60^0$

Hence the triangle is an equilateral triangle.

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

Let $a,b,c$ and $d$ be non-zero numbers. If the point of intersection of the lines $4ax+2ay+c=0$ and $5bx+2by+d=0$ lies in the fourth quadrant and is equidistant from the two axes then

  1. $2bc-3ad=0$
  2. $2bc+3ad=0$
  3. $3bc-2ad=0$
  4. $3bc+2ad=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Since point of intersection lies in the fourth quadrant and is equidistant from coordinate axes,

the $x$ and $y$ co-ordinates will be same 

Hence the coordinates become $(h,-h)$

Passing $4ax+2ay+c=0$ through $(h,-h)$ 

$\Rightarrow 4ah-2ah+c=0$

$\Rightarrow h=-\dfrac{c}{2a}----------(1)$

Also passing the second line $5bx+2by+d=0$ through $(h,-h)$

$\Rightarrow 5bh-2bh+d=0$

$\Rightarrow h=-\dfrac{d}{3b}----(2)$

From eq (1) and (2)

$-\dfrac{c}{2a}=-\dfrac{d}{3b}$

$\Rightarrow 3bc-2ad=0$
Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

The straight line passes through the point of intersection of the straight lines $x+2y-10=0$ and $2x+y+5=0$, is 

  1. $5x-4y=0$
  2. $5x+4y=0$
  3. $4x-5y=0$
  4. $4x+5y=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The line passes through the point of intersection of the equations $x+2y-10=0$  and $2x+y+5=0$

Now,
$\ x+2y-10=0....(i)\ 2x+y+5=0....(ii)\times 2\ =>4x+2y+10=0....(iii)$
Subtracting (iii) and (i), we get,
$-3x-20=0\ =>x=\cfrac { -20 }{ 3 } \ \therefore y=\cfrac { 25 }{ 3 } $
Now the line must pass through $(\cfrac { -20 }{ 3 } ,\cfrac { 25 }{ 3 } )$ 
Therefore, $5x+4y=0$ passes through $(\cfrac { -20 }{ 3 } ,\cfrac { 25 }{ 3 } )$

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

If the line $y-\sqrt{3}x+3=0$ cuts the curve $y^{2}=x+2$ at $A$ and $B$ and point on the line $P$ is $\left(\sqrt{3},0\right)$ then $\left|PA.PB\right|=$

  1. $\dfrac{4\left(\sqrt{3}+2\right)}{3}$
  2. $\dfrac{4\left(2-\sqrt{3}\right)}{3}$
  3. $\dfrac{4\sqrt{3}+}{2}$
  4. $\dfrac{2\left(\sqrt{3}+2\right)}{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Line $ = y - \sqrt 3 x + 3 = 0$
$\begin{array}{l} y=\sqrt { 3 } x-3 \\ m=\sqrt { 3 } =\tan  \theta  \\ \theta =\frac { \pi  }{ 3 } =60^{ \circ  }\to \left( i \right)  \end{array}$
Parametric form of line
$\begin{array}{l} =\frac { { x-{ x _{ 1 } } } }{ { \cos  \theta  } } =\frac { { y-{ y _{ 1 } } } }{ { \sin  \theta  } } =r  \\ \therefore x={ x _{ 1 } }+r \cos  \theta  \\ y={ y _{ 1 } }+r \sin  \theta  \end{array}$
As we know $P\left( {\sqrt 3 ,0} \right)$
$\therefore \left. \begin{array}{l} x=\sqrt { 3 } +r \cos  \theta  \\ y=0+r \sin  \theta  \end{array} \right\} lie\, \, on\, \, parabola\, \, { y^{ 2 } }=x+2$
$\begin{array}{l} \therefore { \left( { r\sin  \theta  } \right) ^{ 2 } }=\sqrt { 3 } +r\cos  \theta +2 \\ { r^{ 2 } }{ \sin ^{ 2 }  }\theta =\sqrt { 3 } +r\cos  \theta +2 \\ { r^{ 2 } }{ \sin ^{ 2 }  }\theta -r\cos  \theta -\sqrt { 3 } -2=0\, \, \, \begin{array} { *{ 20 }{ c } }{ { r _{ 1 } }=PA } \\ { { r _{ 2 } }=PB } \end{array} \\ { r _{ 1 } }{ r _{ 2 } }=\frac { { -\sqrt { 3 } -2 } }{ { { { \sin   }^{ 2 } }\theta  } } \to \left( { ii } \right)  \\ from\, \, \left( i \right) \, \, \theta =60^{ \circ  }\, \, \sin  60^{ \circ  }=\frac { { \sqrt { 3 }  } }{ 2 } \, \, { \sin ^{ 2 }  }60=\frac { 3 }{ 4 }  \\ from\, \, \left( { ii } \right) \, \, \frac { { -\sqrt { 3 } -2 } }{ { \frac { 3 }{ 4 }  } } =\frac { { -4\left( { \sqrt { 3 } +2 } \right)  } }{ 3 }  \\ \therefore \left| { PA.PB } \right| =\left| { { r _{ 1 } }{ r _{ 2 } } } \right| =\frac { { 4\left( { \sqrt { 3 } +2 } \right)  } }{ 3 }  \end{array}$
Hence ans is A.