Tag: mathematics and statistics

Quizzes (38) Questions (955)

Questions Related to mathematics and statistics

Multiple choice mathematics and statistics inverse of a matrix and linear equations application of determinants area of triangle and collinearity of three points applications of determinants

Let O(0, 0), P(3,4), Q(6, 0) be the vertices of the triangle OPQ. The point R inside the triangle OPQ is such that the triangles OPR,PQR, OQR are of equal area. The coordinates of R are 

  1. $\displaystyle \left ( \frac{4}{3}, 3 \right )$
  2. $\displaystyle \left ( 3, \frac{2}{3} \right )$
  3. $\displaystyle \left ( 3, \frac{4}{3} \right )$
  4. $\displaystyle \left ( \frac{4}{3}, \frac{2}{3} \right )$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let coordinate of $R = (a,b)$

Given area of triangle OPR, PQR, and OQR are same. 

$\cfrac{1}{2}\begin{vmatrix} 0&0&1\3&4&1\a&b&1\end{vmatrix}=\cfrac{1}{2}\begin{vmatrix} 3&4&1\6&0&1\a&b&1\end{vmatrix}=\cfrac{1}{2}\begin{vmatrix} 0&0&1\6&0&1\a&b&1\end{vmatrix}$

$\Rightarrow 3b-4a=24-4a-3b=6b$.

Solving this equation be get $a=3, b =\cfrac{4}{3}$

Multiple choice mathematics and statistics inverse of a matrix and linear equations application of determinants area of triangle and collinearity of three points applications of determinants

The co-ordinates of the vertices A, B, C of a triangle are $ \displaystyle \left ( 6,3 \right ),\left ( -3,5 \right ),\left ( 4,-2 \right ) $ respectively and P is any point $ \displaystyle \left ( x,y \right ), $ then the ratio of areas of triangles PBC and ABC is

  1. $ \displaystyle \begin{vmatrix}x-y-2\end{vmatrix}:7 $
  2. $ \displaystyle \begin{vmatrix}x+y+2\end{vmatrix}:7 $
  3. $ \displaystyle \begin{vmatrix}x+y-2\end{vmatrix}:7 $

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let  $ P=(x,y)$

We have area of $\displaystyle \triangle PBC=\left| \frac { 1 }{ 2 } \begin{vmatrix} x\quad  & y\quad  & 1 \\ -3 & 5 & 1 \\ 4 & -2 & 1 \end{vmatrix} \right| $

$\displaystyle =\frac { 1 }{ 2 } \left| \left[ x\left( 5+2 \right) -3\left( -2-y \right) +4\left( y-5 \right)  \right]  \right| $

$\displaystyle =\frac { 1 }{ 2 } \left| 7x+7y-14 \right| =\frac { 7 }{ 2 } \left| x+y-2 \right| $

Area of $\displaystyle \triangle ABC=\left| \frac { 1 }{ 2 } \begin{vmatrix} 6\quad  & 3\quad  & 1 \\ -3 & 5 & 1 \\ 4 & -2 & 1 \end{vmatrix} \right| $

$\displaystyle =\frac { 1 }{ 2 } \left| \left[ 6\left( 5+2 \right) -3\left( -2-3 \right) +4\left( 3-5 \right)  \right]  \right| $

$\displaystyle \\ =\dfrac { 1 }{ 2 } \left| 42+15-8 \right| =\dfrac { 49 }{ 2 } $

$\displaystyle \therefore \frac { area\triangle PBC }{ area\triangle ABC } =\dfrac { \left| x+y-2 \right|  }{ 7 } $
Multiple choice mathematics and statistics inverse of a matrix and linear equations application of determinants area of triangle and collinearity of three points applications of determinants

if $ \displaystyle a,b,c $ as well as $ \displaystyle d,e,f $ are in G.P. with same common ratio then set of points $ \displaystyle \left ( a,d \right ),\left ( b,e \right ),\left ( c,f \right ) $ are

  1. collinear

  2. concurrent

  3. lies on a circle

  4. lie on an ellipse
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Are of triangle formed by the given points is,
$\Delta  = \cfrac{1}{2}\left|\begin{vmatrix}a&d&1\b&e&1\c&f&1\end{vmatrix}\right|$
Let common ratio is $r$
$\Rightarrow \Delta = \cfrac{1}{2}\left|\begin{vmatrix}a&d&1\ar&dr&1\ar^2&dr^2&1\end{vmatrix}\right|$
taking $a$ and $d$ common from first and second column respectively,
$\Delta =  \cfrac{ad}{2}\left|\begin{vmatrix}1&1&1\r&r&1\r^2&r^2&1\end{vmatrix}\right| = 0$, Since first and second column are same.
Hence given points are collinear.

Multiple choice mathematics and statistics inverse of a matrix and linear equations application of determinants area of triangle and collinearity of three points applications of determinants

The vertices of the triangle $ABC$ are $(2, 1, 1), (3, 1, 2), (-4, 0, 1)$. The area of triangle is

  1. $\displaystyle \frac{3\sqrt{38}}{2}$
  2. $\sqrt{38}$
  3. $\displaystyle \frac{\sqrt{38}}{2}$
  4. $4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The vertices of the triangle $ABC$ are $(2,1,1),(3,1,2),(-4,0,1)$
$\overrightarrow { AB } =i+k$ and $\overrightarrow { AC } =-6i-j$
now, $\displaystyle \triangle =\frac { \left| \overrightarrow { AB } \times \overrightarrow { AC }  \right|  }{ 2 } =\frac { \left| \left( i+k \right) \times \left( -6i-j \right)  \right|  }{ 2 } =\frac { \left| i-6j-k \right|  }{ 2 } $
Therefore, $\triangle =\dfrac { \sqrt { 38 }  }{ 2 } $

Ans: C

Multiple choice mathematics and statistics inverse of a matrix and linear equations application of determinants area of triangle and collinearity of three points applications of determinants

Let $\displaystyle A\left ( x _{1},y _{1} \right ),B\left ( x _{2},y _{2} \right ), C\left ( x _{3},y _{3} \right )$ be three points. Area of triangle with vertices $A, B,C$ is given by
$\displaystyle \frac{1}{2}\left | \Delta  \right |$ where,  

$\displaystyle \Delta =\begin{vmatrix}x _{1} &y _{1}  &1 \ x _{2} & y _{2}  & 1\ x _{3} &y _{3}  &1 \end{vmatrix}$.

If $\displaystyle a=BC,b=CA,c=AB$ and $\displaystyle 2s=a+b+c$, then $\displaystyle \Delta ^{2}$ equals

  1. $\displaystyle abc $
  2. $\displaystyle s(s-a)(s-b)(s-c)$
  3. $\cfrac {abc}{4} $
  4. $\displaystyle 4s(s-a)(s-b)(s-c)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given area$=\cfrac { 1 }{ 2 } \left| \triangle  \right| $

Heron's formula

Area$=\sqrt { S\left( S-a \right) \left( S-b \right) \left( S-c \right)  } $

$\cfrac { 1 }{ 2 } \left| \triangle  \right| =\sqrt { S\left( S-a \right) \left( S-b \right) \left( S-c \right)  } $

Squaring on both sides

$=\cfrac { { \left| \triangle  \right|  }^{ 2 } }{ 4 } =S\left( S-a \right) \left( S-b \right) \left( S-c \right) $

${ \left| \triangle  \right|  }^{ 2 }=4S\left( S-a \right) \left( S-b \right) \left( S-c \right) $

Option D
Multiple choice mathematics and statistics inverse of a matrix and linear equations application of determinants area of triangle and collinearity of three points applications of determinants

Let $\displaystyle A\left ( x _{1},y _{1} \right ),B\left ( x _{2},y _{2} \right ), C\left ( x _{3},y _{3} \right )$ be three points. Area of triangle with vertices $A, B,C$ is given by $\displaystyle \frac{1}{2}\left | \Delta  \right |$ where,  $\displaystyle \Delta =\begin{vmatrix}x _{1} &y _{1}  &1 \\
x _{2} & y _{2}  & 1\\
x _{3} &y _{3}  &1
\end{vmatrix}$.If $\displaystyle \triangle ABC$ is an equilateral triangle and $\displaystyle a = BC$ is a rational number, then $\displaystyle \triangle$ must be

  1. an integer

  2. a rational number

  3. an irrational number

  4. an imaginary number
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If $a$ is rational then $a^{ 2 }$ is also rational 
Now as $\Delta =\dfrac { \sqrt { 3 }  }{ 4 } a^{ 2 }$
Then $\Delta $ is irrational

Multiple choice mathematics and statistics inverse of a matrix and linear equations application of determinants area of triangle and collinearity of three points applications of determinants

What is the area of the triangle formed by the points $(a,c+a), (a,c)$ and $(-a,c-a)$?

  1. $\displaystyle- a^{2}$
  2. $\displaystyle \frac{1}{a^{2}}$
  3. $\displaystyle a^{2}+a$

  4. zero
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\left( a,c+a \right)  \left( a,c \right)  \left( -a,c-a \right) $

$\triangle \begin{vmatrix} 1 & 1 & 1 \ a & a & -a \ c+a & \quad c & \quad c-a \end{vmatrix}$
$ac-{ a }^{ 2 }+ac-(ac-{ a }^{ 2 }+ac+{ a }^{ 2 })+ac-ac-{ a }^{ 2 }$
$-2{ a }^{ 2 }-2ac+2ac$
$-2{ a }^{ 2 }$
Area $=\cfrac { 1 }{ 2 } \left[ \triangle  \right] =\cfrac { 1 }{ 2 } \left( -{ a }^{ 2 } \right) $
$=-{ a }^{ 2 }$

Multiple choice mathematics and statistics inverse of a matrix and linear equations application of determinants area of triangle and collinearity of three points applications of determinants

What is the area of the triangle formed by the points $(a,c+a), \displaystyle \left ( a^{2},c^{2} \right )$ and $(-a, c-a)$?

  1. $1$
  2. $\displaystyle a^{2}$
  3. $\displaystyle \sqrt{a^{2}+c^{2}}$

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

$(a,c+a)\quad ({ a }^{ 2 },{ c }^{ 2 })\quad (-a,\quad c-a)$

$\triangle =\begin{vmatrix} 1 & 1 & 1 \ a & { a }^{ 2 } & -a \ c+a & { \quad c }^{ 2 } & \quad c-a \end{vmatrix}$
${ a }^{ 2 }c-{ a }^{ 3 }+a{ c }^{ 2 }-ac+{ a }^{ 2 }-ac-{ a }^{ 2 }+a{ c }^{ 2 }+{ a }^{ 2 }c-{ a }^{ 3 }$
$2a{ c }^{ 2 }-2{ a }^{ 3 }-2ac$
Area $=\cfrac { 1 }{ 2 } \triangle $
$=a{ c }^{ 2 }-{ a }^{ 3 }-ac$
OPTION-D

Multiple choice mathematics and statistics inverse of a matrix and linear equations application of determinants area of triangle and collinearity of three points applications of determinants

What is the area of the triangle formed by the points $(a,b+c), (b,c+a)$ and $(c,a+b)$?

  1. $1$
  2. $-1$
  3. $0$
  4. $\displaystyle \frac{1}{2}\left ( abc \right )^{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$\left( a,b+c \right) ;\left( b,c+a \right) ;\left( c,a+b \right) $
$\triangle =\begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ b+c & c+a & a+b \end{vmatrix}$
$ab+{ b }^{ 2 }-{ c }^{ 2 }-ac-{ a }^{ 2 }-ab+bc+{ c }^{ 2 }+ac+{ c }^{ 2 }-{ a }^{ 2 }-bc$
$=0$
Area of triangle$=\cfrac { 1 }{ 2 } \left| \triangle  \right| =0$
Option C
Multiple choice mathematics and statistics inverse of a matrix and linear equations application of determinants area of triangle and collinearity of three points applications of determinants

The area of a triangle whose vertices are (-2,-2), (-1,-3) and (p,0) is 3 sq.units what is the value of p?

  1. -2

  2. 2

  3. 3

  4. -3
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the vertices of the triangle A(-2,-2),B(-1,-3) and C(P,0) then

$Area  of   \triangle ABC=\frac{1}{2}\left[x _1(y _2-y _3)+x _2(y _3-y _1)+x _3(y _1-y _2)\right]$
Area =3 sq. unit
Here$x _1=-2,y _1=-2$
        $x _2=-1,y _2=-3$
         $x _3=p,y _3=0$
$\Rightarrow 3=\frac{1}{2}\left[(-2(-3-0)+-1(0-(-2))+p(-2-(-3)\right]$
$\Rightarrow 3=\frac{1}{2}[6-2+p]$
$\Rightarrow 6=4+p$
$\Rightarrow -p=4-6$
$\Rightarrow p=2$