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The area of a triangle is =54 square units</h2><h2>
The perpendicular distance from B to AC is = ![\frac{108}{\sqrt{149} } units](https://tex.z-dn.net/?f=%5Cfrac%7B108%7D%7B%5Csqrt%7B149%7D%20%7D%20units)
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Step-by-step explanation:
Given a triangle ABC with vertices A(2,1),B(12,2) and C(12,8)
![x_1=2,y_1=1,x_2=12,y_2=2,x_3=12 and y_3=8](https://tex.z-dn.net/?f=x_1%3D2%2Cy_1%3D1%2Cx_2%3D12%2Cy_2%3D2%2Cx_3%3D12%20and%20%20%20%20%20%20y_3%3D8)
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 = ![\sqrt{(x_1-x_3)^{2} +(y_1-y_3)^2}](https://tex.z-dn.net/?f=%5Csqrt%7B%28x_1-x_3%29%5E%7B2%7D%20%2B%28y_1-y_3%29%5E2%7D)
= ![\sqrt{(2-12)^{2} +(1-8)^2}](https://tex.z-dn.net/?f=%5Csqrt%7B%282-12%29%5E%7B2%7D%20%2B%281-8%29%5E2%7D)
=
units
Let the perpendicular distance from B to AC be = x
According To Problem
![\frac{1}{2} \times x \times \sqrt{149} = 54](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20%20x%20%5Ctimes%20%5Csqrt%7B149%7D%20%3D%2054)
⇔
units
Therefore the perpendicular distance from B to AC is = ![\frac{108}{\sqrt{149} } units](https://tex.z-dn.net/?f=%5Cfrac%7B108%7D%7B%5Csqrt%7B149%7D%20%7D%20units)
Answer:
y = 5
Step-by-step explanation:
y = 0x + b
5 = 0(4) + b
5 = 0 + b
5 = b
Answer:
-7
Step-by-step explanation:
30-16= -5n +3n
14 = -2n
multiplying both by half,u have -
7 = -n
multiply both by -1
7 = n