Tag: area of triangle and collinearity of three points

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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 $\begin{vmatrix} x _1 & y _1 & 1 \ x _2 & y _2 & 1 \ x _3 & y _3 & 1\end{vmatrix}=\begin{vmatrix} a _1 & b _1 & 1\ a _2 & b _2 & 1 \ a _3 & b _3 & 1\end{vmatrix}$, 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.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

  1. $-5$ or $35$
  2. $5$ or $-35$
  3. $15$ or $-5$
  4. $-5$ or $-25$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $-35$
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 $\displaystyle \left | \begin{matrix}x _{1} &y _{1}  &1 \ x _{2} &y _{2}  &1 \ x _{3} &y _{3}  &1 \end{matrix} \right |=\left | \begin{matrix}1 &1  &1 \ b _{1} &b _{2}  &b _{3} \ a _{1} &a _{2}  &a _{3}\end{matrix} \right |$ 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

  1. congruent

  2. similar

  3. equal in area

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
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| $

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

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