Question

The vertices of a triangle are A (2, 3), B (5, 6), and C (6, -1). Find the area of AABC. sq. units​

Answer 1

Answer:

Area of triangle

= 1/2 √x1 (y2 - y3) + x2 (y3 - y1) + x3 (y1 - y2)

Here,

• x1 = 2
• x2 = 5
• x3 = 6
• y1 = 3
• y2 = 6
• y3 = -1

Put all value in formula we get,

Area of triangle =

= 1/2 √ 2 ( 6- (-1) ) + 5 ( -1 -3 ) + 6 ( 3 - 6 )

= 1/2 √ 2 × 7 + 5 × (-4) + 6 × (-3)

= 1/2 √ 14 - 20 - 18

= 1/2 √ -24

Here, negative square root is not possible.

So, area of triangle is not possible

Step-by-step explanation:

r

Answer 2

Answer :

x1=2; x2=5;x3=6

y1=3;y2=6;y3=-1

area of the given triangle = 1/2[2(6--1)+5(-1-3)+6(3-6)]

=1/2[14-20-18]

=1/2[14-38]

=1/2[-24]

=-12

area of something can't be in negative therefore

area of the given one is 12 sq. units  

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