Tag: application of determinants

Questions Related to application of determinants

If $\begin{vmatrix} x _1 & y _1 & 1 \ x _2 & y _2 & 1 \ x _3 & y _3 & 1\end{vmatrix}=\begin{vmatrix} a _1 & b _1 & 1\ a _2 & b _2 & 1 \ a _3 & b _3 & 1\end{vmatrix}$, then the two triangles with vertices $(x _1, y _1), (x _2, y _2), (x _3, y _3)$ and $(a _1,b _1)$, $(a _2, b _2)$, $(a _3, b _3)$ must be congruent.

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. True

  2. False

Show answer
Correct Option: B
Explanation:

The two determinants denote twice the area of $\Delta^s$ whose vertices are $(x _1,y _1), (x _2, y _2)$, $(x _3, y _3)$ and $(a _1, b _1)$, $(a _2, b _2), (a _3, b _3)$. This the equality of two determinants implies that their areas are equal. But equality of the areas of two triangles does not imply that they are congruent.

If the area of the triangle with vertices $(2, 5), (7, k)$ and $(3, 1)$ is $10$, then find the value of $k$.

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. $-5$ or $35$

  2. $5$ or $-35$

  3. $15$ or $-5$

  4. $-5$ or $-25$

Show answer
Correct Option: B
Explanation:
If $(x _1,y _1), (x _2, y _2)$ ans $(x _3, y _3)$ are the vertices of a triangle, then its area is given by $\pm \dfrac {1}{2}[x _1(y _2-y _3)+x _2(y _3-y _1)+x _3(y _1-y _2)]$ 
Given vertices are $(2,5), (7,k), (3,1)$ and area is $10$.
Therefore, $\pm 10 = \dfrac {1}{2}[2(k-1)+7(1-5)+3(5-k)]$
$\Rightarrow \pm 20=2k-2-28+15-3k$
$\Rightarrow \pm 20=-k-15$
$\Rightarrow k = 5$ or $-35$

If $\displaystyle \left | \begin{matrix}x _{1} &y _{1}  &1 \ x _{2} &y _{2}  &1 \ x _{3} &y _{3}  &1 \end{matrix} \right |=\left | \begin{matrix}1 &1  &1 \ b _{1} &b _{2}  &b _{3} \ a _{1} &a _{2}  &a _{3}\end{matrix} \right |$ then the two triangles whose vertices are $\displaystyle \left ( x _{1},y _{1} \right ), \left ( x _{2},y _{2} \right ), ( \left ( x _{3},y _{3} \right ) $ and $\displaystyle\left ( a _{1},b _{1} \right ), \left ( a _{2},b _{2} \right ), \left ( a _{13},b _{3} \right ),$ 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. congruent

  2. similar

  3. equal in area

  4. none of these

Show answer
Correct Option: C
Explanation:

If $\left( x _{ 1 },y _{ 1 } \right) ,\left( x _{ 2 },y _{ 2 } \right) ,(\left( x _{ 3 },y _{ 3 } \right) $ are the vertices of triangle , then its area is 

$A _{1}=\dfrac { 1 }{ 2 } \left| \begin{matrix} x _{ 1 } & y _{ 1 } & 1 \ x _{ 2 } & y _{ 2 } & 1 \ x _{ 3 } & y _{ 3 } & 1 \end{matrix} \right| $

If $\left( a _{ 1 },b _{ 1 } \right) ,\left( a _{ 2 },b _{ 2 } \right) ,\left( a _{ 3 },b _{ 3 } \right) $ are the vertices of triangle , then its area is

$A _{2}=\dfrac { 1 }{ 2 } \left| \begin{matrix} a _{ 1 } & b _{ 1 } & 1 \ a _{ 2 } & b _{ 2 } & 1 \ a _{ 3 } & b _{ 3 } & 1 \end{matrix} \right| $

$A _{2}=\dfrac { 1 }{ 2 } \left| \begin{matrix} a _{ 1 } & a _{ 2 } & a _{ 3 } \ b _{ 1 } & b _{ 2 } & b _{ 3 } \ 1 & 1 & 1 \end{matrix} \right|    (\because |A|=|A^{T}|)$


$A _{2}=-\dfrac{1}{2}\left| \begin{matrix} 1 & 1 & 1 \ b _{ 1 } & b _{ 2 } & b _{ 3 } \ a _{ 1 } & a _{ 2 } & a _{ 3 } \end{matrix} \right| $

Since, area is positive,

$A _{2}=\dfrac{1}{2}\left| \begin{matrix} 1 & 1 & 1 \ b _{ 1 } & b _{ 2 } & b _{ 3 } \ a _{ 1 } & a _{ 2 } & a _{ 3 } \end{matrix} \right| $

Given, $\left| \begin{matrix} x _{ 1 } & y _{ 1 } & 1 \ x _{ 2 } & y _{ 2 } & 1 \ x _{ 3 } & y _{ 3 } & 1 \end{matrix} \right| =\left| \begin{matrix} 1 & 1 & 1 \ b _{ 1 } & b _{ 2 } & b _{ 3 } \ a _{ 1 } & a _{ 2 } & a _{ 3 } \end{matrix} \right| $

$\Rightarrow A _{1}=A _{2}$

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) $
$=-{ a }^{ 2 }$