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  • Find the area of the triangle whose vertices are (–8, 4), (–6, 6) and (–3, 9).

Find the area of the triangle whose vertices are (–8, 4), (–6, 6) and (–3, 9).

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best_answer

Solution.
Area \, of \, triangle = \frac{1}{2}\left [ x_{1} \left ( y_{2} -y_{3}\right )+x_{2} \left ( y_{3} -y_{1}\right )+x_{3} \left ( y_{1} -y_{2}\right )\right ]
Given vertices are (–8, 4), (–6, 6), (–3, 9)
x1 =-8    x2 =-6     x3 = -3
y1=4    y2= 6     y3= 9
We know that area of triangle is
= \frac{1}{2}\left [ x_{1} \left ( y_{2} -y_{3}\right )+x_{2} \left ( y_{3} -y_{1}\right )+x_{3} \left ( y_{1} -y_{2}\right )\right ]
= \frac{1}{2}\left [ \left ( -8 \right )\left ( 6-9 \right )+\left ( -6 \right )\left ( 9-4 \right )+\left ( -3 \right ) \left ( 4-6 \right )\right ]
= \frac{1}{2} \left [ 24+30+6 \right ]
= \frac{1}{2} \left [ 60 \right ]
= 30 sq.units

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