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1. Find the area of the triangle with vertices at the points:(iii) (-1,-8), (-2,-3) and (3,2) (iv) (0,0), (6,0) and (4,3)
CBSE | Class 12 | Determinants | Exercise 6.3 | Maths | RD Sharma
1. Find the area of the triangle with vertices at the points:(iii) (-1,-8), (-2,-3) and (3,2) (iv) (0,0), (6,0) and (4,3)

(iii) According to the question (-1,-8), (-2,-3) and (3,2) are the vertices of the triangle.

We know that if vertices of a triangle are \left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right) and \left( {{x}_{3}},{{y}_{3}} \right), then the area of the triangle is given by:

\vartriangle =\frac{1}{2}\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & 1 \\ {{x}_{3}} & {{y}_{3}} & 1 \\ \end{matrix} \right|

Then putting the value in the above formula,

\vartriangle =\frac{1}{2}\left| \begin{matrix} -1 & -8 & 1 \\ -2 & -3 & 1 \\ 3 & 2 & 1 \\ \end{matrix} \right|

Expand it along {{R}_{1}}

=\frac{1}{2}\left[ -1\left| \begin{matrix} -3 & 1 \\ 2 & 1 \\ \end{matrix} \right|-8\left| \begin{matrix} -2 & 1 \\ 3 & 1 \\ \end{matrix} \right|+1\left| \begin{matrix} -2 & -3 \\ 3 & 2 \\ \end{matrix} \right| \right].

=\frac{1}{2}\left[ -1\left( -5 \right)-8\left( -5 \right)+1\left( 5 \right) \right]

=\frac{1}{2}\left[ 5-40+5 \right]

=\frac{-30}{2}

=\frac{1}{2}\left[ -1\left| \begin{matrix} -3 & 1 \\ 2 & 1 \\ \end{matrix} \right|-8\left| \begin{matrix} -2 & 1 \\ 3 & 1 \\ \end{matrix} \right|+1\left| \begin{matrix} -2 & -3 \\ 3 & 2 \\ \end{matrix} \right| \right]

=\frac{1}{2}\left[ -1\left( -5 \right)-8\left( -5 \right)+1\left( 5 \right) \right]

=\frac{1}{2}\left[ 5-40+5 \right]

=\frac{-30}{2}

As the area cannot be negative. Hence, 15 square unit is the area

Therefore area of triangle is 15 square units

(iv) As per the qiuestion (-1,-8), (-2,-3) and (3,2) are the vertices of the triangle.

We know that if vertices of a triangle are \left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right) and \left( {{x}_{3}},{{y}_{3}} \right), then the area of the triangle is given by:

\vartriangle =\frac{1}{2}\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & 1 \\ {{x}_{3}} & {{y}_{3}} & 1 \\ \end{matrix} \right|

Then, putting the value in above formula

\vartriangle =\left| \begin{matrix} 0 & 0 & 1 \\ 6 & 0 & 1 \\ 4 & 3 & 1 \\ \end{matrix} \right|

Expanding along {{R}_{1}}

=\frac{1}{2}\left[ 0\left| \begin{matrix} 0 & 1 \\ 3 & 1 \\ \end{matrix} \right|-0\left| \begin{matrix} 6 & 1 \\ 4 & 1 \\ \end{matrix} \right|+1\left| \begin{matrix} 6 & 0 \\ 4 & 3 \\ \end{matrix} \right| \right]

=\frac{1}{2}\left[ 0-0+1\left( 18 \right) \right]

=\frac{1}{2}\left[ 18 \right]

=9

Hence, the area of the triangle is 9 sq.units

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