Question

23. Find the area of the triangle with vertices (-1, 2), (4,0) and
(3,9) by determinants


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Answer 1

Given :

• vertices of triangle are (-1, 2), (4,0) and (3,9)

To find :

• area of triangle whose vertices are given

Formula used :

Area of triangle whose vertices are (x1,y1) ,(x2,y2) and (x3,y3) :-

$\dfrac{1}{2} \times \begin{vmatrix} \rm x_{1}&\rm y_{1} & {1} \\\rm x_{2} &\rm y_{2} & 1 \\\rm x_{3} &\rm y_{3} &1 \\ \end{vmatrix}$

Solution :

Vertices of triangle are = (-1, 2), (4,0) and (3,9)

Area :-

$\implies\dfrac{1}{2} \times \begin{vmatrix} \rm x_{1}&\rm y_{1} & {1} \\\rm x_{2} &\rm y_{2} & 1 \\\rm x_{3} &\rm y_{3} &1 \\ \end{vmatrix}$

$\implies\dfrac{1}{2} \times \begin{vmatrix} \rm - 1&\rm 2 & {1} \\\rm 4 &\rm 0 & 1 \\\rm 3&\rm 9\: &1 \\ \end{vmatrix}$

Expanding from row 1:-

$\sf \implies\dfrac{1}{2} \times[ - 1(0 - 9) - 2( 4 - 3) + 1(36 - 0)]$

$\sf \implies\dfrac{1}{2} \times[ 9 - 2 + 36]$

$\sf \implies\dfrac{1}{2} \times(45 - 2)$

$\sf \implies\dfrac{1}{2} \times43$

$\sf \implies\dfrac{43}{2}$

$\sf \implies21.5 \: square \: units$

Answer :

• The area of the triangle with vertices (-1, 2), (4,0) and (3,9) = 21.5 square unit