Question

Find the area of a triangle whose vertices are (6,3)(-3,5)(4,-2)

Answer 1

Given:

• Vertices of triangle are (6,3) , (-3,5) , (4, - 2).

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To find:

• Area of triangle?

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Solution:

$\setlength{\unitlength}{1.2cm}\begin{picture}(0,0)\linethickness{0.4mm}\qbezier(1, 0)(1,0)(3,3)\qbezier(5,0)(5,0)(3,3)\qbezier(5,0)(5,0)(1,0)\put(2.7, 3.3){\sf (6,3)}\put(0.5, - 0.4){\sf (-3,5)}\put(4.8, -0.4){\sf (4, - 2)}\end{picture}$

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Here,

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• x₁ = 6
• y₁ = 3
• x₂ = - 3
• y₂ = 5
• x₃ = 4
• y₃ = - 2

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★ Now, Finding area of Triangle,

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We know that,

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$\star\;{\boxed{\sf{\purple{Area\;of\; \triangle = \dfrac{1}{2} \bigg[ 6(5 + 2) + (-3)(-2 - 3) + 4(3 - 5) \bigg]}}}}$

$:\implies\sf \dfrac{1}{2} \bigg[ 6 \times 7 - 3 \times (-5) + 4(-2) \bigg]$

$:\implies\sf \dfrac{1}{2} \bigg[ 42 + 15 - 8 \bigg]$

$:\implies\sf \dfrac{1}{2} [ 49 ]$

$:\implies{\boxed{\sf{\pink{24.5\;sq\;units}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Area\;of\;triangle\;is\; \bf{24.5\;sq\;units}.}}}$

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$\qquad\qquad\boxed{\underline{\underline{\bigstar \: \bf\:More\:to\:know\:\bigstar}}}$

• Distance Formula = $\sf \sqrt{(x_2 - x_1) + (y_2 - y_1)}$

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• Section Formula = $\sf (x,y) = \bigg( \dfrac{m_2 x_1 + m_1 x_2}{m_1 + m_2} \;,\; \dfrac{m_2 y_1 + m_1 y_2}{m_1 + m_2} \bigg)$

Answer 2

$\huge\bold{\underline{Question:}}$

Find the area of a triangle whose vertices are (6,3)(-3,5)(4,-2)

$\huge\bold{\underline{Answer:}}$

Given:

Let A(6, 3), B(-3, 5) and C(4,-2) be the given points.

To find:

Find the area of the triangle

Solution:

We know that,

are of a triangle

$\boxed{\bf{\red{\dfrac{1}{2}[x₁(y₂ - y₃)+x₂(y₃ - y₁)+x₃(y₁ - y₂)]}}}$

Here,

x₁ = 6, y₁ = 3, x₂ = -3, y₂ = 5, x₃ = 4, y₃ = -2

Area of the triangle$\sf{\implies \dfrac{1}{2}[6(5+2)+(-3)(-2-3)+4(3-5)]}$

$\sf{\implies \dfrac{1}{2}[6×7-3×(-5)+4(-2)]}$

$\sf{\implies \dfrac{1}{2}[42+15-8]}$

$\sf{\implies \dfrac{49}{2}}$

$\sf{\implies 24.5\:sq.units}$

Therefore,

Area of the triangle is 24.5 sq.units.

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