Question

find the area of the triangle where vertivecs are;-(2,3) ,(0,-1),(-4,2)

Answer 1

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

Here it is given that,

• Vertices of triangle are A(2,3), B(0,-1) and C(-4,2)

➢ We have to find the area of △

We know,

The formula to find area of triangle is :

$\boxed{\rm \dfrac{1}{2} \times | \ x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) \ |}$

Here,

$: \implies x_1 = 2 \\ :\implies x_2 = 0\\ : \implies x_3 = -4\\ :\implies y_1 = 3\\ :\implies y_2 = -1\\ : \implies y_3 = 2$

Now,

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

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

$\sf \implies{\rm \dfrac{1}{2} \times | \ - 6+0+( - 16)\ |}$

$\sf \implies{\rm \dfrac{1}{2} \times | \ - 6 - 16\ |}$

$\sf \implies{\rm \dfrac{1}{2} \times | \ - 22\ |}$

$\sf \implies{\rm \dfrac{1}{2} \times 22}$

$\bf \implies{\rm 11 \: units ^{2} }$

Therefore,

• Area of triangle = 11 units²