What is the area of the triangle formed by the vertices $(0, 0), (2, 1) and (5, 3) .$

$1$

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$\frac{1}{2}$

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$2$

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$4$

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Solution

The correct option is B
$\frac{1}{2}$

Area of a triangle having co-ordinates as
$(x_{1}, y_{1}), (x_{2}, y_{2}), and, (x_{3}, y_{3})$ is given as

$\frac{1}{2} ∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & 1 x_{2} & y_{2} & 1 x_{3} & y_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣$
The given vertices are (0, 0), (2, 1) and (5, 3)
We can put in the value of coordinates of the vertices in this formula to obtain the area required as below.

$Δ = \frac{1}{2} ∣ ∣ ∣ ∣ \begin{matrix} 0 & 0 & 1 2 & 1 & 1 5 & 3 & 1 \end{matrix} ∣ ∣ ∣ ∣$
$= \frac{1}{2} [0 (1 - 3) + 2 (3 - 0) + 5 (0 - 1)]$
$= \frac{1}{2} sq.units.$

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Similar questions

What is the area of the triangle formed by the vertices $(0, 0), (2, 1) and (5, 3) .$

The area of the triangle which is formed by the vertices ABC is

△_{1}

(a m_{1}^{2}, 2 a m_{1}) (a m_{2}^{2}, 2 a m_{2}), (a m_{3}^{2}, 2 a m_{3})

△_{2}

(a m_{1} m_{2}, a (m_{1} + m_{2})) (a m_{2} m_{3}, a (m_{2} + m_{3}))

(a m_{3} m_{1}, a (m_{3} + m_{1}))

\frac{△_{1}}{△_{2}}

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