Question

Let A(x1, y1), B(x2, y2), C(x3, y3). Then find the area of the following triangles in a plane.

Answer 1

Answer:

$Area \: of \triangle \: ABC\\\\ = \frac{1}{2} \bigg |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\bigg| \\ \:$

Step-by-step explanation:

We know that area of a triangle of given vertex A(x1, y1), B(x2, y2) and C(x3, y3) can be found by the formula given below:

$Area \: of \triangle \: ABC = \frac{1}{2} \bigg|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\bigg| \\ \\ \:$

modulus sign(| |) outside the formula represent that the magnitude only taken up.If sometimes area will come negative,than discard the sign take only absolute value.

Since if area is negative it shows that Triangle does not exist.

Hope it helps you.

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

Here the there vertices of the triangle ABC are

$\sf{A(x_1,y_1), B(x_2,y_2), C(x_3,y_3)}$

The area of the triangle is given by

$= \frac{1}{2} \displaystyle \sf{\begin{vmatrix}x_1 &y_1 & 1\\ x_2&y_2 & 1 \\ x_3 &y_3 & 1 \end{vmatrix} \: \: \: } \: \: sq \: unit$

$= \sf{ \displaystyle \: \frac{1}{2} |x_1(y_2 - y_3) +x_2(y_3 - y_1) +x_3(y_1 - y_2) \: | \: } \: \: sq \: unit$

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