Question

can someone help me In maths assignment please. find the area of a triangle given the co-ordinates of its vertices 1) the respective formula for area of a triangle 2) the graph paper method​

the co-ordinates are (-5,-1),(3,-5),(5,2 )

And please do on graph

I will mark as branilieast

Answer 1

$\tt\large\underline\purple{Let:-}$

$\tt{\implies Coordinates\:for\:(A)=(-5\:\:,-1)}$

$\tt{\implies Coordinates\:for\:(B)=(3\:\:,-5)}$

$\tt{\implies Coordinates\:for\:(C)=(5\:\:,2)}$

$\tt\large\underline\purple{To\:Find:-}$

$\tt{\implies Area\:of\: triangle\:_{(ABC)}=?}$

$\tt\large\underline\purple{Solution:-}$

To calculate the area of triangle at first we have to find out the Area of triangle with the help of Coordinates points or Vertex of traingle. As per the attachment , I assume the triangle ABC with vertex A, B, C and its side are AB , BC and AC. Now calculate it's area by applying formula.

$\tt\large\underline\purple{Formula\:Used:-}$

$\tt{\implies Area=\dfrac{1}{2}[x_1(y_2 - y_3) + x_2(y_3 - y_1)+x_3(y_1 - y_2)]}$

$\tt\large{\underbrace\pink{Answer\implies 32\: units}}$

Answer 2

Given :-

Co-ordinates are (-5,-1), (3,-5), (5,2)

To Find :-

Area using 1) the respective formula for the area of a triangle 2) the graph paper method​

Solution :-

Let

A = (-5,-1) & B = (3,-5) & C = (5,2)

i) Formula

We know that

Area of Δ = 1/2 | x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)

$\bf Here \begin{cases}\sf x_1=\;\&\;\sf y_1=-1 \\\sf x_2=3\;\&\;y_2==-5\\\sf x_3=5\;\&\;y_3=2\end{cases}$

$\sf \implies Area= \dfrac{1}{2}\big| -5(-5 - 2) + 3(2-\{-1\})+5(-1-\{-5\})\big|$

$\sf \implies Area= \dfrac{1}{2}\big|-5 (-7) + 3(2 + 1) + 5(-1 + 5)\bigg|$

$\sf \implies Area= \dfrac{1}{2}\big|-5(-7)+3(3)+5(4)\big|$

$\sf \implies Area= \dfrac{1}{2}\big|35+9+20\big|$

$\sf \implies Area= \dfrac{1}{2}\times 64$

$\sf \implies Area= \dfrac{64}{2}$

$\sf \implies Area= 32\;cm^2$

ii) Graph method

For AB

$\sf D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$\sf D=\sqrt{(3-1)^2+\{3 - (-1)\}^2}$

$\sf D=\sqrt{(2)^2+(3+1)^2}$

$\sf D=\sqrt{4+16}$

$\sf D=\sqrt{20}$

$\sf D=2\sqrt{5}$

For BC

$\sf D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$\sf D=\sqrt{(5-3)^2+\{2-(-5)\}^2}$

$\sf D=\sqrt{(2)^2+(7)^2}$

$\sf D=\sqrt{4+49}$

$\sf D=\sqrt{53}$