With referance to the conventional certesian $(x, y)$ coordinate system, the vertices of the triangle have the following coordinates :

$(x_{1}, y_{1})$ = $(1, 0)$ ; $(x_{2}, y_{2})$ = $(2, 2)$ and $(x_{3}, y_{3})$ = $(4, 3)$ . The area of the triangle is equal to

\frac{3}{2}

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\frac{3}{4}

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\frac{5}{4}

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\frac{5}{2}

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Solution

The correct option is A $\frac{3}{2}$

Area = $\frac{1}{2}$ $∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & 1 x_{2} & y_{2} & 1 x_{3} & y_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣$
= $\frac{1}{2}$ $∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 1 2 & 2 & 1 4 & 3 & 1 \end{matrix} ∣ ∣ ∣ ∣$ = $- \frac{3}{2}$
Since area can not be -ve so answer is $\frac{3}{2}$

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.
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With referance to the conventional certesian (x,y) coordinate system, the vertices of the triangle have the following coordinates : (x1,y1) = (1,0); (x2,y2) = (2,2) and (x3,y3) = (4,3). The area of the triangle is equal to

The correct option is A 32 Area =12∣∣ ∣∣x1y11x2y21x3y31∣∣ ∣∣ =12∣∣ ∣∣101221431∣∣ ∣∣ =−32 Since area can not be -ve so answer is 32

With referance to the conventional certesian $(x, y)$ coordinate system, the vertices of the triangle have the following coordinates :

$(x_{1}, y_{1})$ = $(1, 0)$ ; $(x_{2}, y_{2})$ = $(2, 2)$ and $(x_{3}, y_{3})$ = $(4, 3)$ . The area of the triangle is equal to