Question

Find the area of a triangle whose vertices are (2,- 3), (4,2) and (7,6) respecively.
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Answer 1

Answer:-

Given:

Vertices of a triangle are (2 , - 3) , (4 , 2) & (7 , 6).

We know that,

Area of a triangle with vertices (x₁ , y₁) , (x₂ , y₂) & $\sf (x_3 , y_3)$ is :

$\sf \: \dfrac{1}{2} \begin{vmatrix} \sf \: x _{1} - \sf x _{2} & \sf \: x _{1} - x _{3} \\ \\ \sf \: y _{1} - y _{2}& \sf \: y _{1} - y _{3} \end{vmatrix}$

Let,

• x₁ = 2

• x₂ = 4

• $\sf x_3$ = 7

• y₁ = - 3

• y₂ = 2

• $\sf y_3$ = 6.

Hence,

$\implies \sf Area\: of\: the\: triangle= \frac{1}{2} \begin{vmatrix} \sf 2 - 4 & \sf 2 - 7 \\\\\ \sf - 3 - 2 & \sf -3-6 \end{vmatrix} \\\\\implies\sf Area\: of\: the\: triangle= \frac{1}{2} \begin{vmatrix} \sf - 2 & \sf - 5 \\\\\sf - 5& \sf -9 \end{vmatrix} \\\\ \implies \sf Area \:of \:the \:triangle=\frac{1}{2} \begin{vmatrix} \sf { ( -2)(-9) -( - 5)(-5) }\end{vmatrix} \\\\\implies \sf Area\: of\: the\: triangle = \frac{1}{2} \times 7\\\\\implies \large{\sf Area\:of\:the\: triangle=\frac{7}{2} \:\: unit^2}$

Answer 2

Answer:-

Given:

Vertices of a triangle are (2 , - 3) , (4 , 2) & (7 , 6).

We know that,

Area of a triangle with vertices

$((x₁ , y₁) , (x₂ , y₂) & \\ \sf (x_3 , y_3)(x </p><p>3</p><p> </p><p> ,y </p><p>3</p><p> </p><p> ) is : \\ \\ </p><p></p><p>\begin{gathered}\sf \: \dfrac{1}{2} \begin{vmatrix} \sf \: x _{1} - \sf x _{2} & \sf \: x _{1} - x _{3} \\ \\ \sf \: y _{1} - y _{2}& \sf \: y _{1} - y _{3} \end{vmatrix}\end{gathered} </p><p>$

Let,

$x₁ = 2 \\ </p><p>x₂ = 4 \\ </p><p>\sf x_</p><p>3</p><p> </p><p> = 7$

$y₁ = - 3 \\ </p><p>y₂ = 2 \\ </p><p>\sf y3 = 6</p><p></p><p>$

Hence,

$\begin{gathered}\implies \sf Area\: of\: the\: triangle= \frac{1}{2} \begin{vmatrix} \sf 2 - 4 & \sf 2 - 7 \\\\\ \sf - 3 - 2 & \sf -3-6 \end{vmatrix} \\\\\implies\sf Area\: of\: the\: triangle= \frac{1}{2} \begin{vmatrix} \sf - 2 & \sf - 5 \\\\\sf - 5& \sf -9 \end{vmatrix} \\\\ \implies \sf Area \:of \:the \:triangle=\frac{1}{2} \begin{vmatrix} \sf { ( -2)(-9) -( - 5)(-5) }\end{vmatrix} \\\\\implies \sf Area\: of\: the\: triangle = \frac{1}{2} \times 7\\\\\implies \large{\sf Area\:of\:the\: triangle=\frac{7}{2} \:\: unit^2}\end{gathered}$