Question

(1) Find the area of the triangle whose vertices are (0,0), (1, 2) and (4,3).
​

Answer 1

Answer:

Answer:area of triangle = 2.5 sq m

Answer:area of triangle = 2.5 sq m

Answer:area of triangle = 2.5 sq m

Answer:area of triangle = 2.5 sq m

Answer:area of triangle = 2.5 sq m

Answer 2

Given :

• (0,0), (1, 2) and (4,3) are vertices of a triangle.

To find :

• Area of triangle

Formula used :

Area of triangle :-

$\dfrac{1}{2}\begin{vmatrix} \sf x1& \sf \: y1&\sf \: 1 \\ \\ \sf x2& \sf \: y2& \sf1 \\ \\ \sf \: x3 &\sf \: y3 &\sf 1 \end{vmatrix}$

Where :-

• (x1,y1) , (x2,y2) and ( x3,y3) are vertices of triangle

Solution :

Now , (0,0), (1, 2) and (4,3) are vertices of a triangle.

we will use the following formula to find area of triangle :-

$\dfrac{1}{2}\begin{vmatrix} \sf x1& \sf \: y1&\sf \: 1 \\ \\ \sf x2& \sf \: y2& \sf1 \\ \\ \sf \: x3 &\sf \: y3 &\sf 1 \end{vmatrix}$

Now put :-

• X1 = 0
• X2 = 1
• X3 = 4
• Y1 = 0
• Y2 = 2
• Y3 = 3

$A = \dfrac{1}{2} \begin{vmatrix} \sf 0& \sf \: 0&\sf \: 1\\\\ \sf 1& \sf \: 2& \sf1 \\\\ \sf \: 4 &\sf \: 3 &\sf 1 \end{vmatrix}$

Now expanding from Row 1 :-

⟹ 1/2 | 0( 2-3) -0(1-4)+1(3-8) |

⟹ 1/2 | 0 -0+(3-8) |

⟹ 1/2 | (3-8) |

⟹ 1/2 | -5 |

[ Modulus / Mod is used because area of any figure can never be in negative. It is always positive ]

⟹ 5/2 square unit

Answer :

Area of triangle = 5/2 square unit

____________________

$\large\bf\blue{Additional-Information}$

If there points are collinear then determinant is zero. It is given as :-

$\begin{vmatrix} \sf x1& \sf \: y1&\sf \: 1 \\ \\ \sf x2& \sf \: y2& \sf1 \\ \\ \sf \: x3 &\sf \: y3 &\sf 1 \end{vmatrix}$

where :-

• (x1,y1) , (x2,y2) and ( x3,y3) are collinear points.

___________________