<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 = 
Answer:
x=-17, y=5. (-17, 5).
Step-by-step explanation:
-2x-5y=9
3x+11y=4
---------------
3(-2x-5y)=3(9)
2(3x+11y)=2(4)
---------------------
-6x-15y=27
6x+22y=8
-----------------
7y=35
y=35/7
y=5
3x+11(5)=4
3x+55=4
3x=4-55
3x=-51
x=-51/3
x=-17
7+z / 2
z = 10
replace the value of z
7 +10 /2
use order of operations ( division comes before addition)
7+5 = 12
2(c+7)=2c+14
Use distributive property
2*c +2*7
2c+14=2c+14
Hope it helps!
Most likely 10 because the initial amount is 5