Math, asked by diliptalpada265, 2 days ago

Find the area of ​​the vertex triangle using the determinant:(5,4),(2,5),(2,3).
-------------------------------------------------------
Answer with proper explanation.


Don't spam.


→ Expecting answer from :

★ Moderators
★ Brainly stars and teacher
★ Others best users ​​

Answers

Answered by diwanamrmznu
5

★GIVEN:-

vertex triangle using the determinant:(5,4),(2,5),(2,3).

★FIND

the area of the vertex triangle using the determinant:(5,4),(2,5),(2,3).

EVALUTION★

We know that triangle area determinant formula

 = \frac{1}{2} (  x_{1}( y_{2} -  y_{3}) +  x_{2}( y_{3} -  y_{1}  )   +  x_{3}( y_{1} -  y_{2} ) )

comparision given triangle

  • (5,4),(2,5),(2,3).

( x_{1}. y_{1})( x_{2}. y_{2})( x_{3}. y_{3})

to area

  •  \frac{1}{2}  |5(5 - 3) + 2(3 - 4) + 2(4 - 5)|  \\  \\  =   \frac{1}{2}   |5 \times 2 + 2 \times  - 1 + 2 \times  - 1|  \\  \\   = \frac{1}{2} |10 - 2 - 2|  \\  \\  \frac{1}{2}  |6| \\  \\  \frac{1}{2} \times 6    \\  \\  = 3
  • ===============

area

  • =3 ✅

==================

I hope it helps you

Answered by mathdude500
5

\large\underline{\sf{Solution-}}

We know,

Area of triangle having vertices (a, b), (c, d) and (e, f) respectively is given by

\rm \:  =  \:  \: \dfrac{1}{2}  \: \begin{gathered}\sf \left | \begin{array}{ccc}a&b&1\\c&d&1\\e&f&1\end{array}\right | \end{gathered} \: square \: units

So, According to statement

Given vertices of the triangle are (5, 4), (2, 5), (2, 3)

So, area of required triangle using determinants is given by

\rm \:  =  \:  \: \dfrac{1}{2}  \: \begin{gathered}\sf \left | \begin{array}{ccc}5&4&1\\2&5&1\\2&3&1\end{array}\right | \end{gathered}

\red{\rm :\longmapsto\:\boxed{\tt{  \: OP \: R_1 \:  \to \: R_1 - R_3 \: }}}

and

\red{\rm :\longmapsto\:\boxed{\tt{  \: OP \: R_2 \:  \to \: R_2 - R_3 \: }}}

So, we get

\rm \:  =  \:  \: \dfrac{1}{2}  \: \begin{gathered}\sf \left | \begin{array}{ccc}3&1&0\\0&2&0\\2&3&1\end{array}\right | \end{gathered}

So, expanding along 3rd column, we get

\rm \:  =  \: \dfrac{1}{2}(6 - 0)

\rm \:  =  \: \dfrac{1}{2}(6)

\rm \:  =  \: 3 \: square \: units

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

MORE TO KNOW

1. The determinant value remains unaltered if rows and columns are interchanged.

2. The determinant value is 0, if two rows or columns are identical.

3. The determinant value is multiplied by - 1, if successive rows or columns are interchanged.

4. The determinant value remains unaltered if rows or columns are added or subtracted.

Similar questions