<h2>
The area of a triangle is =54 square units</h2><h2>
The perpendicular distance from B to AC is = 
</h2>
Step-by-step explanation:
Given a triangle ABC with vertices A(2,1),B(12,2) and C(12,8)

The area of a triangle is= ![\frac{1}{2} [x_1(y_2-y_3) +x_2 (y_3- y_1)+x_3(y_1-y_2)]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%5Bx_1%28y_2-y_3%29%20%2Bx_2%20%28y_3-%20y_1%29%2Bx_3%28y_1-y_2%29%5D)
=![|\frac{1}{2} [2(2-8+12(8-1)+12(1-2)]|](https://tex.z-dn.net/?f=%7C%5Cfrac%7B1%7D%7B2%7D%20%5B2%282-8%2B12%288-1%29%2B12%281-2%29%5D%7C)
=
= 54 square units
The length of AC = 
= 
=
units
Let the perpendicular distance from B to AC be = x
According To Problem

⇔
units
Therefore the perpendicular distance from B to AC is = 
<span>
The good answer is 9
(9:9=1 and 27:9=3)</span>
Answer:
I think it's 9 again I think
Step-by-step explanation:
Hi, i hope this helps! The coordinates are (6,6). I figured this out by plotting the given points and remembering that a rectangle has parallel sides and right angles :))
Don't be worried friend :)
-(y + 2) + 8 = 3
=> -(y + 2) = -5
=> y + 2 = 5
=> y = 3