# What is the area of the triangle formed by the origin with the pair of points (4,3) and (2,5) ?

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Let O bet origin whose coordinates are (0,0). Let A(4,3) and B(2,5).

It is required to find the area of the triangle OAB.

We know that the area of the triangle formed by the three points (x,y1), (x2,y2) and (x3 , y3) is given by:

Area = (1/2) |{(x2-x1)(y1+y2) + (x3-x2)(y3+y1)+ (x1-x3)(y1+y2)}|

Substituting the coordinates of O (0 , 0), A (4,3) and B(2,5) in the order in the above formula, we get:

Area of OAB= (1/2)|{(4-0)(0+3) + (2-4)(3+5) +(0-2)(0+5)}|.

Area of OAB= (1/2) |{(12 ) (-2)8 +(-2)5)}|.

Area OAB = (1/2) |{12 -16-10)}|.

Area of OAB = (1/2) |12-26| .

Area of triangle OAB = (1/2)(14) = 7 sq units.

Area of the triangle OAB = 7 sq units.

We'll compute the area of the triangle whose vertices are the origin and the given points, using the determinant of the second order.

If the coordinates of 2 point are: (x1, y1) and (x2, y2), then the area is:

|x1 y1|

A = (1/2) | |

|x2 y2|

A = (1/2)(x1*y2 - x2*y1)

We'll identify the pair (x1,y1) as (4,3) and the pair (x2,y2) as (2,5).

|4 3|

A = (1/2) | |

|2 5|

A = (1/2)(4*5 - 3*2)

A(1/2)(20 - 6)

A = (1/2)(14)

A = 7 square units.

**The area of the given triangle is: A = 7 square units.**