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

1.	Find the area of a triangle whose vertices are (6,3)(-3,5)(4,-2)
Answer» GIVEN: Vertices of TRIANGLE are (6,3) , (-3,5) , (4, - 2). ⠀⠀⠀⠀⠀⠀⠀ To find: AREA of triangle? ⠀⠀⠀⠀⠀⠀⠀ Solution: $\setlength{\unitlength}{1.2cm}\begin{picture}(0,0)\linethickness{0.4mm}\qbezier(1, 0)(1,0)(3,3)\qbezier(5,0)(5,0)(3,3)\qbezier(5,0)(5,0)(1,0)\put(2.7, 3.3){\sf (6,3)}\put(0.5, - 0.4){\sf (-3,5)}\put(4.8, -0.4){\sf (4, - 2)}\end{picture}$ ⠀⠀⠀⠀⠀⠀⠀ Here, ⠀⠀⠀⠀⠀⠀⠀ x₁ = 6 y₁ = 3 x₂ = - 3 y₂ = 5 x₃ = 4 y₃ = - 2 ⠀⠀⠀⠀⠀⠀⠀ ★ Now, Finding area of Triangle, ⠀⠀⠀⠀⠀⠀⠀ We KNOW that, ⠀⠀⠀⠀⠀⠀⠀ $\star\;{\boxed{\sf{\purple{Area\;of\; \triangle = \dfrac{1}{2} \bigg[ 6(5 + 2) + (-3)(-2 - 3) + 4(3 - 5) \bigg]}}}}\\ \\$ $:\implies\sf \dfrac{1}{2} \bigg[ 6 \times 7 - 3 \times (-5) + 4(-2) \bigg]\\ \\$ $:\implies\sf \dfrac{1}{2} \bigg[ 42 + 15 - 8 \bigg]\\ \\$ $:\implies\sf \dfrac{1}{2} [ 49 ]\\ \\$ $:\implies{\boxed{\sf{\pink{24.5\;sq\;units}}}}\;\bigstar\\ \\$ $\therefore\;{\underline{\sf{Area\;of\;triangle\;is\; \bf{24.5\;sq\;units}.}}}$ ⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━ $\qquad\qquad\boxed{\underline{\underline{\bigstar \: \bf\:More\:to\:know\:\bigstar}}} \\ \\$ Distance Formula = $\sf \sqrt{(x_2 - x_1) + (y_2 - y_1)}$ ⠀⠀⠀⠀⠀⠀⠀ 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»

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:

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