Question

2. If A(3, 8), B(4, -2) and C(5, -1) are the vertices of AABC. Then, its area is (a) 28 – sq. units (b) 37 - sq. units , (c) 57 sq. units 1 2 (d) 75 sq. units 3 TL​

Answer 1

Step-by-step explanation:

We have,

Vertices of triangle are $\tt{\blue{A(3,8)},\,\, \blue{B(4,-2)} \&\,\,\blue{C(5,-1)} }$

Now,

$\sf{its \: \: area \: \:}, \triangle = \dfrac{1}{2} \cdot \left| \left| \begin{array}{ccc} 3 & 8 &1 \\ 4 & - 2 &1 \\ 5 & - 1 &1 \end{array} \right| \right|$

$= \dfrac{1}{2} \cdot \left| 3 \left| \begin{array}{cc} - 2 &1 \\ - 1 &1 \end{array} \right| - 8 \left| \begin{array}{cc} 4 &1 \\ 5 &1 \end{array} \right| + 1 \left| \begin{array}{cc} 4 & - 2 \\ 5 & - 1 \end{array} \right| \right|$

$= \dfrac{1}{2} \cdot \left| 3 ( - 2 + 1) - 8 (4 - 5) + 1 ( - 4 + 10)\right|$

$= \dfrac{1}{2} \cdot \left| 3 ( - 1) - 8 ( - 1) + 1 ( 6)\right|$

$= \dfrac{1}{2} \cdot \left| - 3 + 8 + 6\right|$

$= \dfrac{1}{2} \cdot \left| 5 + 6\right|$

$= \dfrac{11}{2}$