8 = 8v - 4 (v + 8)
Distribute the 4 through the parentheses
8 = 8v - 4v - 32
combine like terms
8 = 4v - 32
add 32 to both sides
8 + 32 = 4v -32 + 32
combine like terms
40 = 4v
divide each side by 4
40/10 = v
4 = v
v= 4
Find the horizontal and vertical distance between the points. First, subtract y2 - y1 to find the vertical distance. Then, subtract x2 - x1 to find the horizontal distance.
![\huge \mathfrak \blue{Solution \: - }](https://tex.z-dn.net/?f=%20%5Chuge%20%5Cmathfrak%20%5Cblue%7BSolution%20%5C%3A%20%20-%20%7D)
The formula for area of triangle is -
![\frac{1}{2} \times base \times height](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%20%20%20%5Ctimes%20base%20%5Ctimes%20height)
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So, the area of first triangle =
![\frac{1}{2} \times base \: \times height](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%20%20%5Ctimes%20base%20%5C%3A%20%20%5Ctimes%20height)
![\frac{1}{2} \times 4 \times 3](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%20%20%5Ctimes%204%20%20%5Ctimes%203)
![= \pink{6 \: units}](https://tex.z-dn.net/?f=%20%3D%20%20%5Cpink%7B6%20%20%5C%3A%20units%7D)
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Area of second triangle =
![\frac{1}{2 } \times base \times height](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%20%7D%20%20%5Ctimes%20base%20%5Ctimes%20height)
![\frac{1}{2} \times 4 \times 6](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%20%20%5Ctimes%204%20%5Ctimes%206)
![= \pink {12 \: units}](https://tex.z-dn.net/?f=%20%3D%20%20%5Cpink%20%7B12%20%5C%3A%20%20units%7D)
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Area of second triangle - Area of first triangle
![= 12 - 6](https://tex.z-dn.net/?f=%20%3D%2012%20-%206)
![= 6](https://tex.z-dn.net/?f=%20%20%3D%206)
![\sf \underline\purple{The \: area \: increases \: by \: 6 \: square \: units.}](https://tex.z-dn.net/?f=%20%5Csf%20%5Cunderline%5Cpurple%7BThe%20%5C%3A%20area%20%5C%3A%20increases%20%5C%3A%20by%20%5C%3A%206%20%5C%3A%20square%20%5C%3A%20units.%7D)
Answer:
![y-\frac{1}{3}=\frac{3}{4}(x-4)](https://tex.z-dn.net/?f=y-%5Cfrac%7B1%7D%7B3%7D%3D%5Cfrac%7B3%7D%7B4%7D%28x-4%29)
Step-by-step explanation:
we know that
The equation of the line into point slope form is equal to
![y-y1=m(x-x1)](https://tex.z-dn.net/?f=y-y1%3Dm%28x-x1%29)
In this problem we have
![point\ (4,\frac{1}{3})](https://tex.z-dn.net/?f=point%5C%20%284%2C%5Cfrac%7B1%7D%7B3%7D%29)
![m=\frac{3}{4}](https://tex.z-dn.net/?f=m%3D%5Cfrac%7B3%7D%7B4%7D)
substitute the given values
![y-\frac{1}{3}=\frac{3}{4}(x-4)](https://tex.z-dn.net/?f=y-%5Cfrac%7B1%7D%7B3%7D%3D%5Cfrac%7B3%7D%7B4%7D%28x-4%29)
y minus StartFraction one-third EndFraction equals StartFraction 3 Over 4 EndFraction left-parenthesis x minus 4 right-parenthesis.(x – 4)