Question

1. Find the area of the triangle whose vertices are
(1) (2,3) (-1,0), (2, -4)
​

Answer 1

Given :

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

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 = (2,3) (-1,0), (2, -4)

Area :-

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

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

Expanding from row 1:-

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

$\sf \implies\dfrac{1}{2} \times[ 8 + 9 + 4]$

$\sf \implies\dfrac{1}{2} \times21$

$\sf \implies\dfrac{21}{2} \: \: square \: unit$

$\sf \implies 10.5 \: \: square \: unit$

Answer :

the area of the triangle whose vertices are (2,3) (-1,0), (2, -4) = 10.5 square unit