Question

Q18. The area of triangle AOB having vertices A (2,5) 0 (-2,2) and B (4,2)
(a) 12 sq. units
(b) 9 sq. units
units
(d) 24 sq. units
(c) 18 sq.
​

Answer 1

$\boxed{\color{Purple} ANSWER:}$

Let's assume,

$x_{1}=2$

$x_{2}=(-2)$

$x_{3}=4$

$y_{1}=5$

$y_{2}=2$

$y_{3}=2$

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So,

By using,

$\frac{1}{2}$[$x_{1}$($y_{2}$ - $y_{3}$)+[$x_{2}$($y_{3}$ - $y_{1}$)+[$x_{3}$($y_{1}$ - $y_{2}$)]

We get,

⇒$\frac{1}{2}$[2(2-2)+(-2)(2-5)+4(5-2)

⇒$\frac{1}{2}$[2(0)-2 (-3)+4 (3)]

⇒$\frac{1}{2}$[0+6+12]

⇒$\frac{1}{2}$[18]

➪9 sq.units.

Hence the required answer is,

$\boxed{\color{DarkGreen} 9 \: sq. \: units}$

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