Tag: applications of determinants

Questions Related to 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 

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

  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. $\displaystyle \left ( \frac{4}{3}, \frac{2}{3} \right )$

Show answer
Correct Option: C
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 $a=3, b =\cfrac{4}{3}$

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

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

  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

Show answer
Correct Option: C
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 } $

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

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

  1. collinear

  2. concurrent

  3. lies on a circle

  4. lie on an ellipse

Show answer
Correct Option: A
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.

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

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

  1. $\displaystyle \frac{3\sqrt{38}}{2}$

  2. $\sqrt{38}$

  3. $\displaystyle \frac{\sqrt{38}}{2}$

  4. $4$

Show answer
Correct Option: C
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

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

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

  1. $\displaystyle abc $

  2. $\displaystyle s(s-a)(s-b)(s-c)$

  3. $\cfrac {abc}{4} $

  4. $\displaystyle 4s(s-a)(s-b)(s-c)$

Show answer
Correct Option: D
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

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

  1. an integer

  2. a rational number

  3. an irrational number

  4. an imaginary number

Show answer
Correct Option: C
Explanation:

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

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

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

  1. $\displaystyle- a^{2}$

  2. $\displaystyle \frac{1}{a^{2}}$

  3. $\displaystyle a^{2}+a$

  4. zero

Show answer
Correct Option: A
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) $
$=-{ a }^{ 2 }$

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

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

  1. $1$

  2. $\displaystyle a^{2}$

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

  4. None of these

Show answer
Correct Option: D
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

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

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

  1. $1$

  2. $-1$

  3. $0$

  4. $\displaystyle \frac{1}{2}\left ( abc \right )^{2}$

Show answer
Correct Option: C
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

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?

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

  1. -2

  2. 2

  3. 3

  4. -3

Show answer
Correct Option: B
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$x _1=-2,y _1=-2$
        $x _2=-1,y _2=-3$
         $x _3=p,y _3=0$
$\Rightarrow 3=\frac{1}{2}\left[(-2(-3-0)+-1(0-(-2))+p(-2-(-3)\right]$
$\Rightarrow 3=\frac{1}{2}[6-2+p]$
$\Rightarrow 6=4+p$
$\Rightarrow -p=4-6$
$\Rightarrow p=2$